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$a_n=(-1)^{k_n}\frac{1}{n}$ where $(k_n-1)^2<n\leq k_n^2$. Is the series $\sum_n a_n$ convergent? I tried with all the classical methods, but they seem to fail, any hint?

EDIT: I had an idea: $\sum a_n=\sum_k (-1)^kb_k$ where $b_k=\sum_{n=(k-1)^2+1}^{k^2}\frac{1}{n}$, I proved that $b_k\rightarrow0$ I want to prove that it's decreasing, so I can apply Leibniz. So I want to prove that $b_{k+1}\leq b_k$, this is false for $k=1$, but it's true for $k=2,3,4$, so I suppose that the sequence of the $b_k$'s is decreasing if $k>1$, but I need some help to prove it.

EDIT EDIT: I checked with my pc and it's true that $b_{k+1}\leq b_k$ for $k>1$, but I still don't know how to prove it, any hint?

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  • $\begingroup$ I think you mean $(k-1)^2 < n \le k^2$, not $(k_n-1)^2 < n \le k_n^2$. $\endgroup$ – Robert Israel Feb 9 '12 at 4:01
  • $\begingroup$ I wrote $k_n$ because $k$ depends on $n$, is it wrong? $\endgroup$ – John Feb 9 '12 at 4:16
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Write $$b_k=\sum\limits_{n=1}^{2k-1}\frac1{(k-1)^2+n},\qquad b_{k+1}=c_k+\sum\limits_{n=1}^{2k-1}\frac1{k^2+n}, $$ with $$ c_k=\frac1{k^2+2k}+\frac1{k^2+2k+1}. $$ Then $b_k\gt b_{k+1}$ if and only if $d_k\gt c_k$ with $$ d_k=\sum\limits_{n=1}^{2k-1}\left(\frac1{(k-1)^2+n}-\frac1{k^2+n}\right)= \sum\limits_{n=1}^{2k-1}\frac{2k-1}{(k^2+n)((k-1)^2+n)}. $$ For every $1\leqslant n\leqslant 2k-1$, $k^2+n\leqslant k^2+2k$ and $(k-1)^2+n\leqslant k^2$ hence $$ d_k\geqslant\frac{(2k-1)^2}{k^3(k+2)}. $$ On the other hand $k^2+2k+1\geqslant k^2+2k$ hence $$ c_k\leqslant\frac2{k(k+2)}. $$ One sees that $c_k\lt d_k$ as soon as $$ \frac2{k(k+2)}\lt\frac{(2k-1)^2}{k^3(k+2)}, $$ that is, $2k^2\lt(2k-1)^2$, which is equivalent to $2k^2-4k+1\gt0$, which holds for every $k\geqslant2$. Thus, $b_k\gt b_{k+1}$ for every $k\geqslant2$.

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Write it as follows (suppressing superfluous symbols). Note that squares are doubly used as indices, so I'm adding a compensatory $\mathcal{O}(1)$ into the fray to keep things formal: $$\mathcal{O}(1)+\left(\sum_1^4-\sum_4^9\right)+\left(\sum_9^{16}-\sum_{16}^{25}\right)+\cdots +\left(\sum_{(2m-1)^2}^{(2m)^2}-\sum_{(2m)^2}^{(2m+1)^2}\right)+\cdots$$

Now use the asymptotic $H_n \sim \log n+\gamma+\frac{1}{2n}+\mathcal{O}(n^{-2})$ on the individual terms:

$$\sum_{(2m-1)^2}^{(2m)^2}-\sum_{(2m)^2}^{(2m+1)^2}=2\log\left(\frac{2m}{2m-1}\right)-2\log\left(\frac{2m+1}{2m}\right) +\mathcal{O}\left(\frac{1}{m^2}\right).$$

Add into the mix $\log(1+x)=x+\mathcal{O}(x^2)=\log\left(\frac{1}{1-x}\right)$ and we have $\square =\mathcal{O}(m^{-2})$, which shows that this series converges. This may or may not help you, depending on what you can use in your HW.

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  • $\begingroup$ I cannot use the fact that $H_n$~ $log\;n+\gamma+1/2n+O(n^{-2})$, there is another way to prove that the series converges? $\endgroup$ – John Feb 9 '12 at 1:06

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