If $\{j_n\}$ is increasing and $\lim\limits_{n→∞}\frac{j_n}n=+∞$, then $\sum\limits_{n=1}^∞\frac{(-1)^{[\sqrt n]}}{j_n}$ converges My friend gave me an interesting analysis problem, and I was tackling it but it was very hard indeed. Here is the problem:

Let $j_1,j_2,j_3,\cdots$ be a sequence of strictly increasing positive integers such that $\lim\limits_{n\rightarrow \infty} \dfrac{j_n}{n} = +\infty$. Prove that $\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{j_n}$ converges.
  Note: $[x]$ is the greatest integer less than or equal to $x.$ 

I think this is some hardcore analysis problem, I would really love to know how you solve this problem, and possibly a full out solution. I am trying my hardest to solve it too, and would appreciate any help offered. 
My attempt at proof of the above:
We know that $$\lim_{n \rightarrow \infty}\frac{\frac{1}{j_n\sqrt{n}}}{\frac{1}{n\sqrt{n}}}=0$$ and clearly the summation:
$$\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}}$$
converges, so we cann directly deduce that 
$\sum_{n=1}^{\infty} \frac{1}{j_n\sqrt{n}}$ converges by the limit comparison test. Now, let $k_n =\frac{1}{j_n}$, and claim that $S_l=\sum\limits_{n=1}^{l}(-1)^{[\sqrt{n}]}k_n$ is Cauchy. Now we take the terms $S_l-S_{m-1}=\sum\limits_{n=m}^{l}(-1)^{[\sqrt{n}]}k_n$ in to consideration, and to bound the term from above, we delete all terms with $[\sqrt{n}]=[\sqrt{m}]$ if $[\sqrt{m}]$ is odd and insert all missing terms with $[\sqrt{m}] =  [\sqrt{n}]$ if  $[\sqrt{m}]$ is even. Hence for a fixed $l,$ taking $1< m < l$, there exists $n,s$ with $n\le s$ such that:
$$S_l-S_{m-1}=\sum\limits_{t=m}^{l}(-1)^{[\sqrt{t}]}k_t\le \sum\limits_{t=(2n)^2}^{(2s+1)^2-1}(-1)^{[\sqrt{t}]}k_t\le $$ $$\sum\limits_{t=(2n)^2}^{(2s+1)^2-1}k_t + \sum^s_{t=n+1}\bigg[2k_{(2t+1)^2-2}+\sum_{u=0}^{(2t)^2-(2t-1)^2-1}(k_{(2t)^2+u}-k_{(2t-1)^2=u})\bigg]$$
$$\le (4n+1)k_{(2n)^2}+\sum^{s}_{t=n+1}2k_{(2t+1)^2-2}\le (4n+1)k_{(2n)^2}+2\sum^s_{t=n+1}\frac{1}{8t}\sum^{(2t+1)^2-2}_{u=(2t-1)^2-1}k_u$$
$$\le (4n+1)k_{(2n)^2}+\sum^s_{t=n+1}\sum^{(2t+1)^2-2}_{u=(2t-1)^2-1}\frac{k_u}{\sqrt{u}}\le  (4n+1)k_{(2n)^2}+\sum^{(2s+1)^2-2}_{u=(2n+1)^2-1}\frac{k_u}{\sqrt{u}} $$
It is not hard to see that as $m \rightarrow \infty,$ $n \rightarrow \infty$. We finally claim that the last expression goes to $0$ as $n \rightarrow \infty$. Note that the second term of the last expression tends to $0$ as as $n \rightarrow \infty$ as $\frac{k_u}{\sqrt{u}}$ converges. So it remains to show that the first term tends to $0$. We prove this by contradiction.
Proof: For some $\epsilon > 0,$ assume that $(4n+1)k_{(2n)^2} >\epsilon$ infinitely often. If we recursively take $n_t>2n_{t-1}$ for $t \ge 1,$ with $(4n+1)k_{(2n)^2}>\epsilon$, then:
$$\sum_{l=1}^{\infty} \frac{k_l}{\sqrt{l}}\ge \sum_{t=1}^{\infty}\sum_{u=n_t^2}^{(2n_t)^2}\frac{k_u}{\sqrt{u}}\le \sum_{t=1}^{\infty}3n_t^2 \frac{k_{(2n_t)^2}}{2n_t}=\frac{3}{2}\sum^{\infty}_{t=1}n_t k_{(2n_t)^2}$$ $$\ge \frac{3}{10}\sum^{\infty}_{t=1}(k_{(2n_t)^2} \ge \frac{3}{10}\sum^{\infty}_{u=1}\epsilon =+\infty$$
This is a clear contradiction. So $\lim_{l\rightarrow \infty}\sup\{S_l-S_{m-1}:m<l\}\le 0$. We can repeat the above with $S_l-S_{m-1}$ bounded from below and get $\lim_{l\rightarrow \infty}\inf\{S_l-S_{m-1}:m<l\}\ge 0$, hence clearly $ S_l$ is Cauchy and thus converges. 
The reason I tried to come with a different idea is that I do not know what Dedekind's Test is, and even though I looked it up just so I can understand the accepted answer, it did not seem to make me fully satisfied. Although I am more aware of Dirichlet's Criterion for series. But of course I now come to realize that they are infact similar.
 A: It seems to me that we don't need the assumption that $j_n/n\to\infty$; I believe that the following is a proof, and unless I missed it I never used $j_n/n\to\infty$:
Say $$I_k=\{n:k^2\le n<(k+1)^2\}=\{n:[\sqrt n]=k\}.$$Let $$a_k=\sum_{n\in I_k}\frac1{j_n}.$$Since $j_n\ge n$ it's clear that $a_k\to0$.
I've been stuck on showing that $a_k$ is decreasing. If so then $\sum(-1)^ka_k$ converges, and it's not quite trivial but not hard to deduce from that that the original sum converges.
Wait - it's enough to show that the sequence $a_k$ has bounded variation.
Each term in $a_k$ can be matched with a corresponding smaller term in the definition of $a_{k+1}$: If $j\in I_k$ then $j'=j+2k+1\in I_{k+1}$, and $1/n_j\ge 1/n_{j'}$. There are two terms in $a_{k+1}$ left over; since $j_n\ge n$ it follows that $$a_k-a_{k+1}\ge-\frac c{k^2}.$$
So $\sum(a_k-a_{k+1})^-$ converges; since $\sum (a_k-a_{k+1})$ converges and $|t|=t+2t^-$ this shows $\sum|a_k-a_{k+1}|$ converges, qed.
Details: First, why do $a_k\to0$ and $\sum|a_k-a_{k+1}|<\infty$ imply that $\sum(-1)^ka_k$ converges? This is a simple generalization of the alternating series test, proved by summation by parts. In fact it's Someone's Test... (Or maybe not. I thought it was Dirichlet's Test, but looking it up that requires $a_{k+1}\le a_k$. In any case it follows easily from summation by parts. Or just note this: Let $$\sigma_K=\sum_{k=1}^K(-1)^ka_k.$$Then $\sigma_{2K}$ is a partial sum of the absolutely convergent series $\sum(a_{2k}-a_{2k-1})$, and $\sigma_{2K+1}-\sigma_{2K}\to0$.
Edit: Daniel Fischer tells us it's Dedekind's Test. Unfortunately the statement there is wrong. The actual DT is this: If the partial sums of $\sum e_j$ are bounded, $b_j\to0$ and $\sum|b_{j+1}-b_j|<\infty$ then the series $\sum e_j b_j$ converges. The statement of Dedekind's Test at EOM omits the hypothesis that $b_j\to0$; taking $b_j=1$ and $e_j=(-1)^j$ shows that this  is nonsense.)
Second, why does convergence of one series imply convergence of the other?  Let $\sigma_K$ be as above and let $$s_N=\sum_{n=1}^N\frac{(-1)^{[\sqrt n]}}{j_n}.$$Then $$\sigma_K=s_{N_K},$$where $N_K=(K+1)^2-1$, so convergence of $s_N$ implies convergence of $\sigma_K$. Conversely, if $N_K<N<N_{K+1}$ then we have either $\sigma_K\le s_N\le\sigma_{K+1}$ or $\sigma_{K+1}\le s_N\le\sigma_K$, depending on whether $K$  is even or odd; hence convergence of $\sigma_K$ implies convergence of $s_N$.
But it's false if we don't assume the $j_n$ are increasing - I mention this because it's already typed up, I missed that hypothesis at first and posted a wrong counterexample. Define $$j_n=\begin{cases} k^2\log(k),&((2k)^2\le n<(2k+1)^2),
\\2k^2\log(k),&((2k+1)^2\le n<(2k+2)^2).\end{cases}$$
(No, $j_n$ is not an integer. If you really think that might matter take $j_n$ to be the integer part of what's defined above.)
Then if you look carefully at your series it seems to me it diverges, because the series $$\sum_k\sum_{n=(2k)^2}^{(2k+2)^2-1}\frac{(-1)^{[\sqrt n]}}{j_n}\sim\sum\left(\frac1{2k\log(k)}-\frac1{k\log(k)}\right)$$
diverges...
