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Assume that $(a_n)_n$ is a sequence of positive real number such that: $$ \sum_{n=1}^{\infty} \frac{a_n}{2n-1}= 1$$ Then show that $$ \sum_{k=1}^{\infty}\sum_{n=1}^{k} \frac{a_n}{k^2}\le 2$$

My Attempt By Fubuni I have, $$ \sum_{k=1}^{\infty}\sum_{n=1}^{k} \frac{a_n}{k^2} = \sum_{n=1}^{\infty}a_n\sum_{k=n}^{\infty} \frac{1}{k^2}$$

Then it suffices to prove that $$\sum_{k=n}^{\infty} \frac{1}{k^2}\le\frac{2}{2n-1}$$ to conclude.

Can someone help ?

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Change the summation limits: $$\begin{cases}1\le k \le +\infty, \\1\le n \le k\end{cases} \quad \text{is equivalent to}\quad \begin{cases} n\le k\le +\infty\\ 1\le n\le +\infty\end{cases}$$ Interchanging the order of summation is not a problem (why?), so \begin{align}\sum_{k=1}^{\infty}\sum_{n=1}^k \frac{a_n}{k^2}&=\sum_{n=1}^{\infty}\left(a_n\sum_{k=n}^{\infty}\frac{1}{k^2}\right)\\[0.2cm]&\le \sum_{n=1}^{\infty}\left(a_n\int_{n-\frac12}^{\infty}\frac{1}{k^2}dk\right)=\sum_{n=1}^{\infty}a_n\left[-\frac{1}{k}\right]_{n-\frac12}^{\infty}\\[0.3cm]&=\sum_{n=1}^{\infty}\frac{a_n}{n-\frac12}=2\sum_{n=1}^{\infty}\frac{a_n}{2n-1}=2\cdot1=2\end{align}

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  • $\begingroup$ Why is the integral great than the sum? $\endgroup$
    – Guy Fsone
    Commented Nov 1, 2017 at 12:55
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    $\begingroup$ You may think of the integral as $$\sum_{k=n, \text{with step dx}}^{\infty}\frac{1}{k^2}\ge \sum_{k=n, \text{with step 1}}^{\infty}\frac{1}{k^2}$$ $\endgroup$
    – Jimmy R.
    Commented Nov 1, 2017 at 12:56
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    $\begingroup$ We have $$\int_n^{\infty} \frac{dt}{t^2} < \sum_{k = n}^{\infty} \frac{1}{k^2}$$ since $$\int_m^{m+1} \frac{dt}{t^2} = \frac{1}{m} - \frac{1}{m+1} = \frac{1}{m(m+1)} < \frac{1}{m^2}.$$ But by convexity, $$\sum_{k = n}^{\infty} \frac{1}{k^2} < \int_{n - \frac{1}{2}}^{\infty} \frac{dt}{t^2}.$$ $\endgroup$
    – Daniel Fischer
    Commented Nov 1, 2017 at 13:06
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    $\begingroup$ @ΒασίληςΜάρκος That's the upper Darboux sum. The lower Darboux sum would be $$\sum_{k = n+1}^m \frac{1}{k^2}.$$ $\endgroup$
    – Daniel Fischer
    Commented Nov 1, 2017 at 13:12
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    $\begingroup$ @GuyFsone For a convex $f$ and $a < b$, we have $$\int_a^b f(x)\,dx \geqslant f\biggl(\frac{a+b}{2}\biggr)\cdot (b-a),$$ with the inequality strict if $f$ is strictly convex. $\endgroup$
    – Daniel Fischer
    Commented Nov 1, 2017 at 13:29

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