trying to solve the nested sum $\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2} $ This sum grabbed my curiousity
$$
\sum_{n=1}^{\infty}  \frac{1}{1^2+2^2+\dots+n^2}
$$
I want to solve it with Calc II methods (that's the math I have).
By getting a closed form for the expression in the denominator, it becomes this:
$$
6 \sum_{n=1}^{\infty}  \frac{1}{n(n+1)(2n+1)} 
$$
First thing I did was partial fractions
$$
6 \sum_{n=1}^{\infty} \left( \frac{1}{n}  + \frac{1}{n+1}-  \frac{4}{2n+1} \right)
$$
You can reindex the first two terms, but I couldn't reindex the $2n+1$ term to telescope with that combined term, my thought is it's not possible because it isn't in the form (n+integer).
So I tried expressing it as an integral, viewing each term as multiplied by an $x$ evaluated between 0 and 1. Then deriving with respect to $x$ to produce an integral.
$$
6 \sum_{n=1}^{\infty}   \int_0^1  (x^{n-1} +x^n  - 4x^{2n}) dx
$$
Switching the order of integration and summation, and doing some reindexing:
$$
6 \int_0^1  \left(3+ 2\sum_{n=0}^\infty x^n - 4 \sum_{n=0}^\infty (x^2)^n\right)dx
$$
I've been doing decimal approximations the whole time, checking that my steps work. They do up to and including that point, the answer is about $1.3$
Next I used the geometric series expansion, and this is where the approximation goes wrong, so I believe this is a problematic step:
$$
6 \int_0^1  \left(3+  \frac{2}{1-x} - \frac{4}{1-x^2} \right)dx
$$
This integral is easy to solve using partial fractions on the $x^2$ term. You get an answer of $6(3-2\ln2)$, which is about $9.6$
The correct answer, based on some website that doesn't show steps, is $6(3-4\ln2)$, which matches my approximation.
I'm really stumped on what I'm doing wrong and how to get that correct answer. I've checked my algebra over and over and learned some latex to write this post. Any help is appreciated!

We have
\begin{align}
\sum_{n=1}^\infty\left(\frac{1}{n}+\frac{1}{n+1}-\frac{4}{2n+1}\right) & =
\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{2}{2n+1}+\frac{1}{n+1}-\frac{2}{2n+1}\right)\\
&= \sum_{n=1}^\infty\left(\frac{1}{n}-\frac{2}{2n+1}\right)+\sum_{n=1}^\infty\left(\frac{1}{n+1}-\frac{2}{2n+1}\right)\\
&= 2\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2n+1}\right)+2\sum_{n=1}^\infty\left(\frac{1}{2n+2}-\frac{1}{2n+1}\right)\\
&= 2\;(1-\log 2)+2\;(\tfrac{1}{2}-\log 2)\\
&=3-4\log 2
\end{align}
where we used $\displaystyle\log 2 = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n} = \sum_{n=1}^\infty\left(\frac{1}{2n-1}-\frac{1}{2n}\right)$.
