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Determine $N$ of series $\sum^N_\limits{n=1}(-1)^n\frac{1}{2n+1}$ so that it differs from $\sum\limits^\infty_{n=1}(-1)^n\frac{1}{2n+1}$ by less than $\frac1{1000}$

How do I do this problem?

I know that the remainder is equal to the sum to infinity - the sum to N should be less than 0.001 given by the formula

$$|R_n|=|S-S_n| \le b_{n+1}$$ $$\int^\infty_1\frac1{2x+1}-\int_N^\infty\frac1{2x+1} \le 0.001$$ $$\infty - \frac12\left[\ln(2n+1)\right]^\infty_N \le 0.001$$

Which doesn't really make any sense... at least to me

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    $\begingroup$ In an alternating series, the error made by truncating at a certain place has absolute value less than the absolute value of the first "neglected" term. $\endgroup$ – André Nicolas Feb 21 '16 at 19:23
  • $\begingroup$ @AndréNicolas And in this case this is essentially sharp, by looking at sums of consecutive pairs of terms. $\endgroup$ – Igor Rivin Feb 21 '16 at 19:27
  • $\begingroup$ See the Hint in my answer. $\endgroup$ – Mhenni Benghorbal Feb 21 '16 at 19:33
  • $\begingroup$ is it just me or is stackexchange being super slow $\endgroup$ – Panthy Feb 21 '16 at 19:33
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    $\begingroup$ You should deal with the inequality $a_{n+1} = \frac{1}{2n+3} < 0.001$ $\endgroup$ – Mhenni Benghorbal Feb 21 '16 at 19:59
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Let $$ R_N = \sum_{n\ge N+1} (-)^n{1\over 2n+1} $$ observe that $$ |R_N| < {1\over 2N+1} $$ then we would like $$ \eqalign{ & R_N < {1\over 1000} \cr \rightarrow\ & {1\over 2N+1} < {1\over 1000} \cr \rightarrow\ & 2N+1 > 1000 \cr \rightarrow\ & N \ge 500 \cr } $$

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  • $\begingroup$ makes sense... thank you $\endgroup$ – Panthy Feb 21 '16 at 22:09
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$$ \begin{align} \sum_{k=2n}^\infty(-1)^k\frac1{2k+1} &=\sum_{k=n}^\infty\left(\frac1{4k+1}-\frac1{4k+3}\right)\\ &=\sum_{k=n}^\infty\frac2{(4k+1)(4k+3)}\\ &\lt\sum_{k=n}^\infty\frac2{4k(4k+4)}\\ &=\frac12\sum_{k=n}^\infty\left(\frac1{4k}-\frac1{4k+4}\right)\\ &=\frac1{8n} \end{align} $$ Thus, at $n=125$, the error is less than $\frac1{1000}$.


$$ \begin{align} \sum_{k=2n-1}^\infty(-1)^{k-1}\frac1{2k+1} &=\sum_{k=n}^\infty\left(\frac1{4k-1}-\frac1{4k+1}\right)\\ &=\sum_{k=n}^\infty\frac2{(4k-1)(4k+1)}\\ &\gt\sum_{k=n}^\infty\frac2{(4k-1)(4k+3)}\\ &=\frac12\sum_{k=n}^\infty\left(\frac1{4k-1}-\frac1{4k+3}\right)\\ &=\frac1{8n-2} \end{align} $$ Thus, at $n=125$, the error is greater than $\frac1{1000}$.


Thus, the first partial sum to be within $\frac1{1000}$ of the infinite sum is $$ \sum_{k=1}^{249}(-1)^k\frac1{2k+1} $$

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