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 ?

• You messed up at one of the steps, so I reccomend comparing them
– user253055
Sep 6, 2015 at 16:39
• @AaronThompson What do you mean? Sep 6, 2015 at 16:41
• @AaronThompson $\int_0^1x^{n-1}=\int_0^12x^{2n-1}=\frac{1}{n}$ Sep 6, 2015 at 16:44
• I am saying he made a mistake in one of the calculations and was doing it right in the other calculations. It's not like he's doing it two different ways
– user253055
Sep 6, 2015 at 16:44
• @AaronThompson (S)he is doing it in multiple ways Sep 6, 2015 at 16:46

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 $\int_{0}^{1} 2x^{2n-1} \,dx$.

• Thank you. But why the second solution is true? Sep 6, 2015 at 17:17
• I have known that uniformly convergent series can be interchanged the order of integration and summation. 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 ? Sep 6, 2015 at 17:18
• @Bless, That's right. We have $$\left| \frac{1}{1+x} - \sum_{k=1}^{n} (2x^{2k-2} - x^{k-1}) \right| = \frac{x^n (1+x-2x^n)}{1-x^2},$$ and as $x \to 1$ the right hand side converges to $n-\frac{1}{2}$. So the first one does not converge uniformly. Sep 6, 2015 at 17:23
• Of course, failure of uniform convergence does not necessarily prevent us from interchanging the order, but in our case this certainly does. Sep 6, 2015 at 17:29
• @Bless, the second one does not converge uniformly, neither. An easy way to see is that, while the partial sum $\sum_{k=1}^{n} (2x^{2k-2} - 2x^{2k-1})$ is zero at $x = 1$ the limiting function $1/(x+1)$ is not zero at $x = 1$. But in this case we have at least two different lifesavers: Abel's theorem and dominated convergence theorem. Either way can be used to justify your second argument. Sep 6, 2015 at 17:37