Inequality used in the proof of Kolmogorov Strong Law of Large Numbers I'm trying show that  convergence follows.
$$\sum_{k \geq 1} \frac{\sigma^2_k}{k^2} < \infty \Rightarrow \lim_{M \rightarrow \infty}\frac{1}{M^2}\sum_{k \leq M} \sigma^2_k=0.$$
Let's consider $D_k = \sum_{n \geq k} \displaystyle\frac{\sigma_n^2}{n^2}$ para $k \geq  1$ and it is noted that for $k=1$ we have to:
$$D_1 = \sum_{n \geq 1} \frac{\sigma_n^2}{n^2} < \infty \text{(by hypothesis)}.$$
or this I have proved that $\lim_{k \rightarrow \infty} D_k = \lim_{k \rightarrow \infty} \sum_{n \geq k} \frac{\sigma_n^2}{n^2} =  0$,
Now note the following:
\begin{eqnarray*}
D_2 &=& \sum_{n \geq 2} \frac{\sigma_n^2}{n^2} = \frac{\sigma_2^2}{2^2} + \frac{\sigma_3^2}{3^2} + \frac{\sigma_4^2}{4^2} \ldots\nonumber \\
D_3 &=& \sum_{n \geq 3} \frac{\sigma_n^2}{n^2} = \frac{\sigma_3^2}{3^2} + \frac{\sigma_4^2}{4^2} \ldots \nonumber\\
D_4 &=& \sum_{n \geq 4} \frac{\sigma_n^2}{n^2} =  \frac{\sigma_4^2}{4^2} \ldots \nonumber\\
&\vdots& \\
\lim_{k \rightarrow \infty} D_k &=& \sum_{n \geq k} \frac{\sigma_n^2}{n^2} =  0. \nonumber
\end{eqnarray*}
then I considered a $M$ such that $M \geq 1$ and I then come to the following equation
$$\frac{1}{M^2}\sum_{k=1}^M \sigma^2_k = \frac{1}{M^2} \sum_{k=1}^M  k^2\left(D_k - D_{k+1}\right).$$
I noted in a book which obtained the following inequality
$$\sum_{n \geq k}  \displaystyle\frac{\sigma^2_k}{k^2} = \frac{1}{M^2} \sum_{k=1}^M  k^2 \left(D_k - D_{k+1}\right)\text{(How could justify this inequality?)} \leq  \frac{1}{M^2} \sum_{k=1}^M  (2k - 1)D_k$$
It is easy to see that
\begin{eqnarray}
\sum_{n \geq k} \displaystyle\frac{\sigma^2_k}{k^2} &=& D_k - D_{k+1}\nonumber  \\
&=& \sum_{n \geq k} \displaystyle\frac{\sigma_n^2}{n^2} - \sum_{n \geq k+1} \displaystyle\frac{\sigma_n^2}{n^2} \nonumber \\
&=& \left\{\displaystyle\frac{\sigma_k^2}{k^2} + \displaystyle\frac{\sigma_{(k+1)}^2}{(k+1)^2} + \ldots \right\} - \left\{\displaystyle\frac{\sigma_{(k+1)}^2}{(k+1)^2} + \displaystyle\frac{\sigma_{(k+2)}^2}{(k+1)^2} + \ldots \right\}.
\end{eqnarray}
where 
$$\lim_{k \rightarrow \infty} \frac{1}{M^2} \sum_{k=1}^M  (2k - 1)D_k = 0$$
but I don´t know to justify that step. Could you please give me a suggestion on how to justify the last step.
Thank you very much, for you help.
 A: \begin{align*}
\sum_{k=1}^M k^2(D_k-D_{k+1})
&= \sum_{k=1}^M k^2 D_k - \sum_{k=1}^M k^2 D_{k+1} \\
&= \sum_{k=1}^M k^2 D_k - \sum_{k=2}^{M+1} (k-1)^2 D_k \\
&= D_1 + \sum_{k=2}^M (k^2-(k-1)^2) D_k - M^2 D_{M+1} \\
&= D_1 + \sum_{k=2}^M (2k-1) D_k - M^2 D_{M+1} \\
&= \sum_{k=1}^M (2k-1) D_k - M^2 D_{M+1} \\
&\le \sum_{k=1}^M (2k-1) D_k
\end{align*}
A: \begin{align*}
\frac{1}{M^2}\sum_{k \leq M}  \sigma^2_k &\le \frac{1}{M^2} \sum_{k=1}^M (2k-1) D_k \\ 
&=\frac{1}{M^2}  \left(D_1 + 3D_2 + 5D_3 + 7D_4   + \ldots \right)\\
&=\frac{1}{M^2} \left(\sum_{n \geq 1}\frac{\sigma^2_n}{n^2} + 3\sum_{n \geq 2}\frac{\sigma^2_n}{n^2} + 5\sum_{n \geq 3}\frac{\sigma^2_n}{n^2}+7\sum_{n \geq 4}\frac{\sigma^2_n}{n^2} + \ldots\right)\\ 
&=\frac{1}{M^2} \left\{\left(\frac{\sigma_1^2}{1^2} + \frac{\sigma_2^2}{2^2} + \ldots \right) + 3\left(\frac{\sigma_2^2}{2^2} + \frac{\sigma_3^2}{3^2} + \ldots \right) + 5\left(\frac{\sigma_3^2}{3^2} + \frac{\sigma_4^2}{4^2} + \ldots \right) + 7\left(\frac{\sigma_4^2}{4^2} + \ldots \right) + \ldots\right\}\\
&=\frac{1}{M^2} \left\{\frac{\sigma_1^2}{1^2} + \frac{4\sigma_2^2}{2^2}  + \frac{9\sigma_3^2}{3^2} + \frac{16\sigma_4^2}{4^2} + \ldots \right\}\\
&= \frac{1}{M^2} \left(\sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2 + \ldots\right)\\
&= \frac{1}{M^2}\sigma_1^2 + \frac{1}{M^2}\sigma_2^2 + \frac{1}{M^2}\sigma_3^2 + \frac{1}{M^2}\sigma_4^2 + \ldots 
\end{align*}
Applying limit
\begin{align*}
\lim_{M \rightarrow \infty} \frac{1}{M^2}\sum_{k=1}^M \sigma^2_k &\le \lim_{M \rightarrow \infty} \left(\frac{1}{M^2}\sigma_1^2 + \frac{1}{M^2}\sigma_2^2 + \frac{1}{M^2}\sigma_3^2 + \frac{1}{M^2}\sigma_4^2 + \ldots \right) \\
&= 0 + 0 + 0 + 0 + \ldots \\
&= 0.
\end{align*}
Therefore
$$\lim_{M \rightarrow \infty} \frac{1}{M^2}\sum_{k \leq M} \sigma^2_k = 0.$$
