Free math worksheets

- Locate the first term after the fourth partial sum. - For an alternating series with decreasing magnitudes, the first omitted magnitude bounds the error.

1. The first term omitted from \(S_4\) is \(\frac15\). 2. The alternating series error bound gives \( |\ln 2-S_4|\leq\frac15\).

\( |\ln 2-S_4|\leq\frac15\).

- Check both conditions on the positive term magnitudes. - The error bound is the magnitude of the first term not included in \(S_4\). - The sign of that first omitted term determines the direction of the error.

1. The positive term magnitudes are \( a_n=\frac{1}{n^2}\). They decrease as \(n\) increases and approach \(0\), so the alternating series error bound applies. 2. The first omitted term has magnitude \(\frac{1}{5^2}\). Therefore, \( |S-S_4|\leq\frac{1}{25}=0.04\). 3. The first omitted term is positive, so the remainder is positive and \(S_4\) is an underestimate.

a) The magnitudes \(\frac{1}{n^2}\) decrease to \(0\). b) \( |S-S_4|\leq\frac{1}{25}=0.04\). c) \(S_4\) is an underestimate.

- Substitute \(x=0.25\) only after selecting the first two terms. - The first omitted term controls the error magnitude. - Use the sign of the omitted term to choose the correct side of the partial sum.

1. The first two nonzero terms give \( S_2=0.25-\frac{0.25^2}{2}=\frac{7}{32}=0.21875\). 2. The first omitted term has magnitude \( \frac{0.25^3}{3}=\frac{1}{192}\approx0.00520833\). 3. The omitted term is positive, so \(S_2\) is an underestimate. Therefore, \( \frac{7}{32}\leq\ln(1.25)\leq\frac{7}{32}+\frac{1}{192}=\frac{43}{192}\).

a) \( S_2=\frac{7}{32}=0.21875\). b) \( |\ln(1.25)-S_2|\leq\frac{1}{192}\approx0.00520833\). c) The partial sum is an underestimate, and \( \frac{7}{32}\leq\ln(1.25)\leq\frac{43}{192}\).

- Choose a partial sum that lies below the exact value. - Track the sign of the first omitted sine term. - Compare the resulting rational lower bound with \(0.84\).

1. The sine series at \(x=1\) is \( \sin 1=1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\cdots\). 2. The fourth partial sum is \( S_4=1-\frac16+\frac{1}{120}-\frac{1}{5040}=\frac{4241}{5040}\approx0.841468\). 3. The next omitted term is positive. Therefore, \(S_4\) is a lower bound for \(\sin 1\). 4. Since \(S_4>0.84\), it follows that \(\sin 1>0.84\).

\( \sin 1\geq\frac{4241}{5040}\approx0.841468>0.84\).

- Substitute \(x=0.5\) into the displayed series. - The first omitted nonzero term gives the error bound. - Use the sign of that omitted term to determine the direction of the error.

1. The first three nonzero terms of the given series give \( S_3=0.5-\frac{0.5^3}{3}+\frac{0.5^5}{10}=0.461458333\). 2. The first omitted term has magnitude \( \frac{0.5^7}{42}\approx0.000186012\). 3. The omitted term is negative, so \(S_3\) is an overestimate.

\(F(0.5)\approx0.461458333\). Error at most \(0.000186012\). The approximation is an overestimate.

- Keep the alternating signs while evaluating the partial sum. - The next geometric term supplies both the bound and the error direction. - Use the infinite geometric sum only after completing the theorem-based estimate.

1. The four-term sum is \( S_4=1-0.6+0.36-0.216=0.544\). 2. The first omitted term is positive with magnitude \(0.6^4=0.1296\). Thus \(S_4\) is an underestimate and \(|S-S_4|\leq0.1296\). 3. The exact sum is \( S=\frac{1}{1+0.6}=0.625\). 4. The exact error is \(0.625-0.544=0.081\), which is smaller than the guaranteed bound \(0.1296\).

a) \(S_4=0.544\). b) Error at most \(0.1296\); \(S_4\) is an underestimate. c) Exact error \(0.081\), which is below the bound.

- Alternating signs are only one condition of the theorem. - Check the limit of the term magnitudes before considering an error bound. - A convergent series must have individual terms that approach \(0\).

1. The term magnitudes are \(b_n=\frac{n}{n+1}\). 2. Although the signs alternate, the magnitudes do not approach \(0\): \( \lim_{n\to\infty}\frac{n}{n+1}=1\). 3. Because the terms of the series do not approach \(0\), the series diverges by the nth-term test. 4. The alternating series theorem does not apply, so there is no valid alternating-series error estimate for its partial sums.

No. The terms do not approach \(0\), so the series diverges and the alternating series error bound is not valid.

- Determine whether the next omitted term raises or lowers the even partial sum. - A rounding guarantee requires every number in the interval to round the same way. - After adding the ninth term, use the magnitude of the tenth term for the new bound.

1. Since the series begins with a positive term, the even partial sum \(S_8\) is an underestimate, and the next term is positive. 2. The alternating series error bound gives \( 1.274\leq S\leq1.274+0.006=1.280\). 3. This interval crosses the rounding boundary \(1.275\), so it does not guarantee a nearest-hundredth value. 4. The ninth partial sum is \( S_9=1.274+0.006=1.280\). It is an overestimate, and the next term is negative with magnitude \(0.002\). 5. Therefore, \( 1.280-0.002\leq S\leq1.280\), so \(S\in[1.278, 1.280]\). 6. Every number in this interval rounds to \(1.28\) to the nearest hundredth.

a) \([1.274, 1.280]\). b) No; the interval crosses \(1.275\). c) \(S_9=1.280\), with guaranteed interval \([1.278, 1.280]\); the guaranteed rounded value is \(1.28\).

- Express the error bound using the index of the first omitted term. - Pay attention to the strict inequality at the tolerance boundary. - Determine the sign of term \(N+1\) after finding \(N\).

1. The first omitted term after \(S_N\) has magnitude \(\frac{1}{(N+1)^3}\). 2. The requirement is \( \frac{1}{(N+1)^3}<0.001=\frac{1}{1000}\). 3. This requires \((N+1)^3>1000\), so \(N+1>10\). The smallest integer choice is \(N=10\). 4. The tenth term is negative and the eleventh term is positive. Therefore, \(S_{10}\) is an underestimate.

a) \(N=10\). b) \(S_{10}\) is an underestimate.

- Evaluate each power of \(1/2\) before combining fractions. - Consecutive alternating partial sums form lower and upper bounds. - The midpoint error cannot exceed half the width of the bracket.

1. Through the second nonzero term, \( S_2=\frac{1}{2}-\frac{(1/2)^3}{3}=\frac{11}{24}\). 2. Through the third nonzero term, \( S_3=S_2+\frac{(1/2)^5}{5}=\frac{223}{480}\). 3. Consecutive partial sums of a decreasing alternating series lie on opposite sides of the exact sum. Thus \( \frac{11}{24}\leq\arctan(0.5)\leq\frac{223}{480}\). 4. The midpoint is \( \frac{1}{2}\left(\frac{11}{24}+\frac{223}{480}\right)=\frac{443}{960}\approx0.461458\). 5. The interval width is \(\frac{1}{160}\), so the midpoint is at most half that distance from the exact value. The guaranteed error is \(\frac{1}{320}=0.003125\).

a) \( S_2=\frac{11}{24}\) and \( S_3=\frac{223}{480}\). b) \( \arctan(0.5)\in\left[\frac{11}{24}, \frac{223}{480}\right]\). c) Midpoint approximation \( \frac{443}{960}\approx0.461458\), with error at most \(0.003125\).

- Use the first omitted term to bound the error after each partial sum. - Compare the entire guaranteed interval with the nearest-thousandth rounding cutoff. - Use the sign of the next omitted term to decide which side of the partial sum contains the exact value.

1. Through the \(x\) term, \(S_1=0.3\). 2. The next-term bound is \( \frac{0.3^3}{3}=0.009\). Since the next term is negative, the exact value lies in \([0.291, 0.3]\). This interval crosses the rounding cutoff \(0.2915\), so it does not guarantee a value rounded to the nearest thousandth. 3. Including the \(x^3\) term gives \( S_2=0.3-\frac{0.3^3}{3}=0.291\). 4. The new error bound is \( \frac{0.3^5}{5}=0.000486\). 5. Since the next term is positive, the exact value lies in \([0.291, 0.291486]\). Every number in that interval rounds to \(0.291\) to the nearest thousandth.

a) \(S_1=0.3\), with error at most \(0.009\). b) No; the guaranteed interval \([0.291, 0.3]\) crosses the rounding cutoff \(0.2915\). c) \(S_2=0.291\), with error at most \(0.000486\), so \(\arctan(0.3)\) rounds to \(0.291\).

- Compare consecutive magnitudes rather than differentiating a continuous extension. - The theorem only requires eventual decrease, not decrease from the first term. - Determine the sign of term \(6\) after finding its magnitude.

1. Let \(a_n=\frac{n^2}{3^n}\). Then \( \frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{3n^2}\). For \(n\geq2\), \((n+1)^2<3n^2\), so the ratio is less than \(1\) and the magnitudes decrease. 2. Removing finitely many initial terms does not affect convergence. From \(n=2\) onward, the series alternates, its magnitudes decrease, and they approach \(0\). 3. The first omitted magnitude after \(S_5\) is \( a_6=\frac{36}{3^6}=\frac{4}{81}\). 4. The sixth term is negative, so the remainder is negative. Thus \(S_5\) is an overestimate and \( |S-S_5|\leq\frac{4}{81}\).

a) \( \frac{a_{n+1}}{a_n}<1\) for \(n\geq2\). b) The tail satisfies the alternating-series conditions. c) \( |S-S_5|\leq\frac{4}{81}\), and \(S_5\) is an overestimate.

- Use the sign of the next term to turn each partial sum into a one-sided interval. - Do not compare only the two partial sums; compare the guaranteed intervals. - The conclusion is certain when the intervals do not overlap.

1. For the logarithm, \( L_3=0.5-\frac{0.5^2}{2}+\frac{0.5^3}{3}=\frac{5}{12}\). The next term is negative with magnitude \(\frac{0.5^4}{4}=\frac{1}{64}\), so \( \ln(1.5)\in\left[\frac{77}{192}, \frac{5}{12}\right]\). 2. For arctangent, \( A_2=0.4-\frac{0.4^3}{3}=\frac{142}{375}\). The next term is positive with magnitude \(\frac{0.4^5}{5}=\frac{32}{15625}\), so \( \arctan(0.4)\in\left[\frac{142}{375}, \frac{17846}{46875}\right]\). 3. The lower bound for \(\ln(1.5)\) is greater than the upper bound for \(\arctan(0.4)\). Therefore, \(\ln(1.5)>\arctan(0.4)\).

\(\ln(1.5)>\arctan(0.4)\).

- Use the first omitted arctangent term as a function of \(x\). - The largest error on \([0, a]\) occurs at the largest input. - Solve the resulting fifth-power inequality for \(a\).

1. The first omitted term has magnitude \(\frac{x^5}{5}\), so \( |\arctan x-P_3(x)|\leq\frac{x^5}{5}\). 2. To guarantee the required error for every \(x\leq a\), require \( \frac{a^5}{5}\leq0.0001\). 3. Thus \(a^5\leq0.0005\), and the largest value is \( a=(0.0005)^{1/5}\approx0.218672\).

\( a=(0.0005)^{1/5}\approx0.218672\).

- Track the denominator of the first term not included after \(N\) terms. - Include the outside factor \(4\) in the error bound. - Check the integer immediately below your proposed answer to prove minimality.

1. After \(N\) nonzero terms, the first omitted denominator is \(2N+1\). 2. Because the entire alternating series is multiplied by \(4\), the error bound is \( \frac{4}{2N+1}\). 3. Require \( \frac{4}{2N+1}<0.01\), so \(2N+1>400\). 4. For \(N=199\), \(2N+1=399\), which is too small. For \(N=200\), \(2N+1=401\), so the bound is below \(0.01\).

\(N=200\) nonzero terms.

- Determine the sign of the next omitted term for each series separately. - A one-sided remainder gives an asymmetric interval around the partial sum. - Add the lower endpoints together and the upper endpoints together.

1. For the logarithm, \( L=0.2-\frac{0.2^2}{2}+\frac{0.2^3}{3}=0.182666667\). The next term is negative with magnitude \(\frac{0.2^4}{4}=0.0004\), so \( \ln(1.2)\in[L-0.0004, L]\). 2. For arctangent, \( T=0.2-\frac{0.2^3}{3}=0.197333333\). The next term is positive with magnitude \(\frac{0.2^5}{5}=0.000064\), so \( \arctan(0.2)\in[T, T+0.000064]\). 3. The combined approximation is \(L+T=0.38\). 4. Adding the interval endpoints gives \( A\in[0.3796, 0.380064]\).

Approximation: \(0.38\). Guaranteed interval: \([0.3796, 0.380064]\).

- Express the first omitted sine term in terms of the number of included terms. - Check one term count that fails and the next one that succeeds. - Use consecutive odd factorials in the approximation.

1. After \(N\) nonzero terms, the first omitted term has magnitude \(\frac{1}{(2N+1)!}\). 2. With \(N=3\), the bound is \(\frac{1}{7!}\approx0.000198413\), which is too large. 3. With \(N=4\), the bound is \(\frac{1}{9!}\approx0.000002756\), which is below \(10^{-5}\). Thus four nonzero terms are required. 4. The approximation is \( 1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}=\frac{4241}{5040}\approx0.841468254\).

a) Four nonzero terms. b) \( \sin 1\approx\frac{4241}{5040}\approx0.841468254\).

- For an alternating series, compare the first omitted term with the target tolerance. - Check the last degree that fails before accepting the next degree. - Use the sign of the first omitted term to determine the direction of the error.

1. The error after the term with \(n=N\) is at most the magnitude of the next term, \(\frac{1}{(N+1)!}\). 2. If \(N=6\), the next term has magnitude \( \frac{1}{7!}\approx0.000198413\), which is too large. 3. If \(N=7\), the next term has magnitude \( \frac{1}{8!}\approx0.0000248016\), which is below \(0.0001\). Thus the least degree is \(7\). 4. The partial sum is \( S_7=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}=\frac{103}{280}\approx0.367857143\). 5. The first omitted term is positive, so \(S_7\) is an underestimate of \(e^{-1}\).

a) Least degree: \(N=7\), so eight terms are included. b) \( e^{-1}\approx\frac{103}{280}\approx0.367857143\), an underestimate, with error at most \( \frac{1}{8!}\approx0.0000248016\).

- Treat the sine and cosine remainders separately before combining them. - The sign of the first omitted term gives the direction of each error. - Use the Pythagorean theorem with the two coordinate-error bounds.

1. For the cosine coordinate, the first omitted term has magnitude \( \frac{0.5^6}{6!}=\frac{1}{46080}\approx0.0000217014\). 2. For the sine coordinate, the first omitted term has magnitude \( \frac{0.5^7}{7!}=\frac{1}{645120}\approx0.00000155010\). 3. In both series, the first omitted term is negative. Therefore, both coordinates of \(q\) are overestimates of the corresponding coordinates of \(u\). 4. The coordinate errors form the legs of a right triangle, so \( \|u-q\|\leq\sqrt{\left(\frac{1}{46080}\right)^2+\left(\frac{1}{645120}\right)^2}\approx0.0000217567\).

a) Cosine-coordinate error at most \( \frac{1}{46080}\approx0.0000217014\); sine-coordinate error at most \( \frac{1}{645120}\approx0.00000155010\). b) Both coordinates of \(q\) are overestimates. c) \( \|u-q\|\leq0.0000217567\).

- Consecutive even and odd partial sums bracket a positive-first alternating series. - Apply an increasing function to both endpoints without reversing the inequalities. - Use the inverse relationship between \(e^x\) and \(\ln x\) only after the interval is established.

1. The fourth partial sum is \( L_4=\frac12-\frac{0.5^2}{2}+\frac{0.5^3}{3}-\frac{0.5^4}{4}=\frac{77}{192}\approx0.401041667\). 2. The fifth partial sum is \( L_5=L_4+\frac{0.5^5}{5}=\frac{391}{960}\approx0.407291667\). 3. For a positive-first alternating series, the even partial sum is below the exact sum and the odd partial sum is above it. Therefore, \( \frac{77}{192}\leq L\leq\frac{391}{960}\). 4. Because \(e^x\) is increasing, \( e^{\frac{77}{192}}\leq e^L\leq e^{\frac{391}{960}}\). 5. Numerically, \( 1.493379491\leq e^L\leq1.502742342\). Since \(e^L=e^{\ln(1.5)}=1.5\), the exact value lies in the guaranteed interval.

a) \( \frac{77}{192}\leq L\leq\frac{391}{960}\). b) \( e^{\frac{77}{192}}\leq e^L\leq e^{\frac{391}{960}}\), or approximately \([1.493379491, 1.502742342]\). c) Yes; \(1.5\) lies in that interval.

- Choose a lower alternating partial sum for \(\sin x/x\). - Choose an upper alternating partial sum for \(\cos x\). - After subtracting, use \(x\leq1\) to prove the polynomial lower bound is positive.

1. For \(0<x\leq1\), the alternating series for \(\sin x/x\) has decreasing term magnitudes. Its two-term partial sum is a lower bound: \( \frac{\sin x}{x}\geq1-\frac{x^2}{6}\). 2. The three-term cosine partial sum is an upper bound: \( \cos x\leq1-\frac{x^2}{2}+\frac{x^4}{24}\). 3. Subtracting gives \( \frac{\sin x}{x}-\cos x\geq\frac{x^2}{3}-\frac{x^4}{24}=\frac{x^2(8-x^2)}{24}\). 4. Since \(0<x\leq1\), \(8-x^2\geq7\). Therefore, \( \frac{\sin x}{x}-\cos x\geq\frac{7x^2}{24}>0\).

\( \frac{\sin x}{x}-\cos x\geq\frac{x^2(8-x^2)}{24}\geq\frac{7x^2}{24}>0\). Therefore, \( \frac{\sin x}{x}>\cos x\) for \(0<x\leq1\).

- Treat each two-term pair as one new term before checking alternation. - The first omitted quantity is a whole block, not one original fraction. - Check the two neighboring block magnitudes to establish the least valid block count. - Use the sign of the first omitted block to determine the direction and interval in part d).

1. Pairing consecutive terms gives \( S=\sum_{k=0}^{\infty}(-1)^kB_k\), where \( B_k=\frac{1}{2k+1}+\frac{1}{2k+2}\). 2. Each \(B_k\) is positive. Both denominators increase with \(k\), so \(B_k\) decreases, and \(B_k\to0\). Thus the grouped series satisfies the alternating-series conditions. 3. If \(K\) complete blocks are retained, the first omitted block is \(B_K\). Near the cutoff, \( B_{19}=\frac{1}{39}+\frac{1}{40}\approx0.05064\) and \( B_{20}=\frac{1}{41}+\frac{1}{42}\approx0.04820\). Therefore, the least number is \(K=20\) blocks, containing \(40\) original terms. 4. The next block has a positive sign, so the \(20\)-block partial sum \(S_{20}\) is an underestimate. Consecutive grouped partial sums give \( S_{20}\leq S\leq S_{20}+B_{20}\).

a) \( S=\sum_{k=0}^{\infty}(-1)^k\left(\frac{1}{2k+1}+\frac{1}{2k+2}\right)\). b) The block magnitudes are positive, decreasing, and approach \(0\). c) \(20\) blocks, or \(40\) original terms. d) The partial sum is an underestimate, and \( S_{20}\leq S\leq S_{20}+\frac{1}{41}+\frac{1}{42}\).

- Rewrite the student's expression as a natural partial sum plus extra positive terms. - Use the bracketing relationship between even and odd natural partial sums. - The first-omitted-term rule depends on sequence order, not the largest denominator shown. - For part d), return to the natural \(20\)-term ordering and use its next term.

1. The alternating-series error bound applies to terms retained in their original order. The quantity \(T\) keeps positive terms through \(\frac1{19}\) but negative terms only through \(\frac1{10}\), so it is not a natural partial sum. 2. Let \(S_N\) be the natural \(N\)-term partial sum. Then \( T=S_{11}+\frac1{13}+\frac1{15}+\frac1{17}+\frac1{19}\). The odd partial sum \(S_{11}\) is an overestimate of \(S\), so \(T>S_{11}>S\). 3. Therefore, \( T-S>T-S_{11}>\frac1{13}>\frac1{20}\). The student's claimed bound is impossible. 4. The natural \(20\)-term partial sum keeps the original order. Its first omitted term has magnitude \(\frac1{21}\), so \( |S-S_{20}|\leq\frac1{21}\).

a) \(T\) skips negative terms while retaining later positive terms, so it is not a natural partial sum. b) \(T>S_{11}>S\), so \(T\) is an overestimate. c) \( T-S>\frac1{13}>\frac1{20}\). d) \( |S-S_{20}|\leq\frac1{21}\).

- Use the parity of the final included term to decide which partial sum is above or below the function. - At the left test value, an upper bound below \(0.4\) is useful; at the right value, use a lower bound above \(0.4\). - A midpoint estimate has error at most half the width of its guaranteed bracket.

1. On \([0, 1]\), the arctangent series is alternating with decreasing term magnitudes. The four-term partial sum ends with a negative term, so it is a lower bound. Adding the next positive term gives an upper bound. 2. At \(x=0.422\), the upper partial sum is approximately \(0.399333<0.4\). Therefore, \(\arctan(0.422)<0.4\). 3. At \(x=0.423\), the lower partial sum is approximately \(0.400133>0.4\). Therefore, \(\arctan(0.423)>0.4\). 4. Since \(\arctan x\) is increasing, \( 0.422<x_*<0.423\). 5. The midpoint is \(0.4225\), and half the interval width is \(0.0005\). Hence \( |x_*-0.4225|<0.0005\).

a) \( x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\leq\arctan x\leq x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}\). b) \(0.422<x_*<0.423\). c) \(x_*\approx0.4225\), with guaranteed error less than \(0.0005\).

- For part a, test the first omitted term after exactly \(40\) nonzero terms. - Apply the tangent subtraction formula before estimating the transformed series. - Compare how quickly powers of \(0.9\) and \(\frac{1}{19}\) decrease.

1. Using \(40\) nonzero terms directly leaves first omitted term \( \frac{0.9^{81}}{81}\approx2.43\times10^{-6}>10^{-6}\). Therefore, \(40\) terms are not enough. 2. The tangent subtraction formula gives \( \tan\left(\frac{\pi}{4}-\arctan(0.9)\right)=\frac{1-0.9}{1+0.9}=\frac{1}{19}\). Both angles are in the principal range, so \( \arctan(0.9)=\frac{\pi}{4}-\arctan\left(\frac{1}{19}\right)\). 3. One term for \(\arctan\left(\frac{1}{19}\right)\) leaves omitted magnitude \( \frac{1}{3\cdot19^3}>10^{-6}\). Two terms leave omitted magnitude \( \frac{1}{5\cdot19^5}\approx8.08\times10^{-8}<10^{-6}\). 4. Therefore, \( \arctan(0.9)\approx\frac{\pi}{4}-\left(\frac{1}{19}-\frac{1}{3\cdot19^3}\right)\), with error at most \(8.08\times10^{-8}\). 5. Powers of \(\frac{1}{19}\) shrink much faster than powers of \(0.9\), so the transformed series reaches the tolerance with far fewer terms.

a) \(40\) direct-series terms are not enough because the first omitted term is approximately \(2.43\times10^{-6}\). b) \( \arctan(0.9)=\frac{\pi}{4}-\arctan\left(\frac{1}{19}\right)\); the transformed series needs \(2\) terms, giving \( \frac{\pi}{4}-\left(\frac{1}{19}-\frac{1}{3\cdot19^3}\right)\) with error at most \(8.08\times10^{-8}\). c) The much smaller input makes successive powers decrease far faster.

- Group the partial sums according to whether the last displayed pair is complete. - For an odd partial sum, write both the partial sum and the first omitted term in terms of the same index \(m\). - Isolate the first omitted term and examine how the remaining tail changes its magnitude. - Compare strict decrease with the repeated equal magnitudes in the displayed series.

1. The positive term magnitudes are \(1,1,\frac12,\frac12,\frac13,\frac13,\ldots\). They are nonincreasing and approach \(0\), which is sufficient for the alternating series theorem. 2. Each even partial sum ends after a complete pair, so \(S_{2m}=0\). Each odd partial sum contains one unpaired positive term, so \(S_{2m-1}=1/m\). Both subsequences approach \(0\). Therefore, the series converges to \(0\). 3. For an odd partial sum, \(|S-S_{2m-1}|=|0-1/m|=1/m\). The first omitted term is \(-1/m\), so its magnitude is also \(1/m\). Equality occurs in the usual alternating-series error bound. 4. In a strictly decreasing alternating series, the terms after the first omitted term begin with an opposite-signed term of smaller magnitude. Their sum cancels a positive amount, but not all, of the first omitted term. Consequently, the remainder has the sign of the first omitted term and magnitude strictly between \(0\) and that term's magnitude. 5. The displayed series has equal consecutive magnitudes, not strictly decreasing magnitudes. Its paired terms can cancel completely, so equality in the bound is possible.

a) The magnitudes are nonincreasing and approach \(0\), so strict decrease is not required. b) The sum is \(0\). c) For every odd partial sum, the absolute error equals the first omitted magnitude: \(|S-S_{2m-1}|=1/m\). d) Strict decrease forces partial, nonzero cancellation after the first omitted term, so the error is strictly smaller. The displayed magnitudes are not strictly decreasing.