<s>[INST] Find the value of $\sum_{n=1}^\infty \frac{1}{n(2n-1)}$ I will show two solutions of this problem.
First solution :
$$\sum_{n=1}^\infty \frac{1}{n(2n-1)}=\sum_{n=1}^\infty \left(\frac{2}{2n-1}-\frac{1}{n}\right)$$
$$=\sum_{n=1}^\infty \left(\int_0^1 (2x^{2n-2}-x^{n-1})dx\right)$$
$$=\int_0^1 \left( \sum_{n=1}^\infty(2x^{2n-2}-x^{n-1})\right)dx$$
$$=\int_0^1 \left( \frac{2}{1-x^2}-\frac{1}{1-x} \right) dx $$
$$=\int_0^1  \frac{1}{1+x}dx $$
$$=\ln 2$$
Second solution :
$$\sum_{n=1}^\infty \frac{1}{n(2n-1)}=\sum_{n=1}^\infty \left(\frac{2}{2n-1}-\frac{1}{n}\right)$$
$$=\sum_{n=1}^\infty \left(\int_0^1 (2x^{2n-2}-2x^{2n-1})dx\right)$$
$$=\int_0^1 \left( \sum_{n=1}^\infty(2x^{2n-2}-2x^{2n-1})\right)dx$$
$$=\int_0^1 \left( \frac{2}{1-x^2}-\frac{2x}{1-x^2} \right) dx $$
$$=\int_0^1  \frac{2}{1+x}dx $$
$$=2\ln 2$$
In fact, $\displaystyle\sum_{n=1}^\infty \frac{1}{n(2n-1)}=2\ln 2$.
Why the first solution is false but the second is true ?
I have known that uniformly convergent series can be integrated term by term.
Is this mean that $\sum_{n=1}^\infty(2x^{2n-2}-2x^{2n-1})$ ; second solution,  is  uniformly convergent but $\sum_{n=1}^\infty(2x^{2n-2}-x^{n-1})$ ; first solution,  is not ?
 [/INST] Here is an even more baffling calculation:
\begin{align*}
0 &= \sum_{n=1}^{\infty} \left( \frac{2}{2n} - \frac{1}{n} \right) \\
&= \sum_{n=1}^{\infty} \int_{0}^{1} (2 x^{2n-1} - x^{n-1}) \, dx \\
&= \int_{0}^{1} \sum_{n=1}^{\infty} (2 x^{2n-1} - x^{n-1}) \, dx \\
&= \int_{0}^{1} \left( \frac{2x}{1-x^2} - \frac{1}{1-x} \right) \, dx \\
&= - \int_{0}^{1} \frac{dx}{x+1} \\
&= -\log 2.
\end{align*}
So where did we mess up? It is the third step, where we interchanged the order of integration and summation. In this step, we are actually dealing with $\infty - \infty$ type indeterminate, which are cleverly hidden, and we failed to manage it properly. This is more evident if we plot the graph of the partial sum
$$y = \sum_{k=1}^{n} (2x^{2k-1} - x^{k-1})$$
for $n = 1, \cdots, 20$ as follows. (Color changes from red to green to blue as $n$ increases.)

The mass of $\log 2$ is concentrated around the high spike at $x = 1$, which vanishes as $n \to \infty$. This is why in both calculation we end up losing $\log 2$.

Speaking differently, the common mistake we have done is essentially the same as in the following bogus argument:
\begin{align*}
0
&= \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \right) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \right) \\
&= \left( \frac{2}{2} + \frac{2}{4} + \frac{2}{6} + \cdots \right) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \right) \\
&= -1 + \left(\frac{2}{2} - \frac{1}{2}\right) - \frac{1}{3} + \left( \frac{2}{4} - \frac{1}{4}\right) - \cdots \\
&= -1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \cdots \\
&= -\log 2.
\end{align*}
In our calculation, shifting the order of the summand is achieved by doubling the exponent; that is, by changing $\int_{0}^{1} x^{n-1} \,dx$ by $\int_{0}^{1} 2x^{2n-1} \,dx$.
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