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I have the following sum:$$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+++++------...$$

How can I get an expression for its partial sum ${s_n}$? I know that I can´t rearrange terms unless it converges absolutely, so which is the best way to approach this kind of series? Any help will be appreciated.

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    $\begingroup$ Hint: Investigate the partial sum of the first $\displaystyle\sum_{k=1}^n=\frac{n(n+1)}2$ terms. $\endgroup$ – Lucian May 21 '15 at 5:22
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What you can do is group together sequences of consecutive terms if they have the same sign.

A typical group of consecutive, equally signed terms is:

$$\pm\sum_{j=t(n)+1}^{t(n+1)}\frac1j$$

where $t(n)$ is the $n$th triangular number, that is, $t(n)=1+2+\cdots+n=\frac{n(n+1)}2$.

and now we can try to bound this sum: $$0<\sum_{j=t(n)+1}^{t(n+1)}\frac1j\le\frac{n+1}{t(n)+1}=\frac{n+1}{1+\frac{n(n+1)}2}\stackrel{n\to\infty}\to0$$

Since it is an alternating series whose terms tend to $0$, the sum converges.

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  • $\begingroup$ How did you find that bound $\endgroup$ – mobzopi May 21 '15 at 7:55
  • $\begingroup$ The sum has $n+1$ terms and every term is lesser or equal that the first. $\endgroup$ – ajotatxe May 21 '15 at 8:04
  • $\begingroup$ Sorry but how can I express a partial sum of the original series? $\endgroup$ – mobzopi May 21 '15 at 8:25

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