My proof that this series sums to log 2,

Question

EDIT: Please see my attempt below at a proof to show the answer is $log2$

Let $$\alpha = \lim_{n \to \infty} \sum_{j=1}^n \frac {(-1)^{j+1}}{j}$$

Part (1): Show that $\alpha$ exists.

Part (2): Show that $\alpha = log2$.

My work:

$\alpha$ exists, by the alternating series test - simple enough. But how can I get started on part(2)?

Proof:

Start with $log(1+x)$. Rewrite it as

$$log(1+x) = \int \frac{1}{1+x}dx$$

Should there be a constant, c, added to the L.H.S.? Or, should I look to make the R.H.S. a definite integral instead?

Continuing along, I get that

$$\int \frac{1}{1+x}dx$$ $$= \int [\sum_{n=0}^{\infty} (-x)^n]dx$$

with the sum valid for $|x|<1$.

Inside the interval of convergence of this sum, i.e., inside (-1,1), we have uniform convergence, and so we can integrate term-by-term, getting

$$ \sum_{n=0}^{\infty} \int (-x)^ndx$$ $$= \sum_{n=0}^{\infty} \frac {(-x)^{n+1}}{n+1}$$

same question: do I add a constant to the result of the indefinite integration?

Now relabeling indices gives

$$ \sum_{n=1}^{\infty} \frac {(-x)^{n}}{n}$$

$$ = \sum_{n=1}^{\infty} \frac {(-1)^{n+1}x^{n}}{n}$$

but we started with $log(1+x)$, so this shows that we have the equation:

$$ log(1+x) = \sum_{n=1}^{\infty} \frac {(-1)^{n+1}x^{n}}{n}$$

and this implies that

$$ log(1+1) = log(2) = \sum_{n=1}^{\infty} \frac {(-1)^{n+1}1^{n}}{n}$$

$$ \ => \alpha = log(2) = \sum_{n=1}^{\infty} \frac {(-1)^{n+1}}{n}$$

as required.

What do you think? Am I close?

My concerns are the indefinite integration that I used a couple of times.

Also, I completely neglected the interval (and boundary) of convergence for $log(1+x)$, and just plugged in $x=1$, since I arrived at an equation. Is this ok to do?

Any hints or comments are welcome.

Thanks,

Answer 1

You can prove that for all $x\in(-1,1)$, $$\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$$ and use Abel theorem.

Answer 2

You need to use Abel partial summation theorem to prove that $$ \log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n $$ is uniform convergent on $[0,1]$ and thus continuous on on $[0,1]$. Hence $$ \lim_{x\to1}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\log{2} $$ Let $B_n=\sum_{k=m}^n \frac{(-1)^{k+1}}{k}$. Since $\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}$ converges, by Cauchy Criterion, for any $\epsilon>0$, there is a $N$ such that $$ |B_n|<\epsilon \quad\text{whenever }\quad n,m>N\tag1 $$ We have \begin{align} \sum_{k=m}^n \frac{(-1)^{k+1}}{k}x^k&=\sum_{k=m}^n (B_k-B_{k-1})x^k \\ &=\sum_{k=m}^n B_kx^k -\sum_{k=m}^n B_{k-1}x^k \\ &=\sum_{k=m}^{n-1} B_k(x^k-x^{k+1})+B_nx^n\tag{2} \end{align} Note: $B_{m−1}=0$.

Since $\lim_{n\to\infty}x^n$ exists and $\:x^n \downarrow$ on $[0,1]$, for all $k>0$ and $x\in[0,1]$ we have $$ x^k-x^{k+1}\geqslant0\: $$ Since for all $k>m$, $-\epsilon<B_k<\epsilon$ $$ |B_k(x^k-x^{k+1})|<\epsilon(x^k-x^{k+1})\tag3 $$ So for all $n,m>N-1$ and $x\in[0,1]$, by $(1)$, $(2)$ and $(3)$ there is \begin{align} \left|\sum_{k=m}^n \frac{(-1)^{k+1}}{k}x^k\right|&\leqslant\sum_{k=m}^{n-1} |B_k(x^k-x^{k+1})|+|B_nx^n| \\ &\leqslant\sum_{k=m}^{n-1} \epsilon\:(x^k-x^{k+1})+\epsilon \:x^n \\ &=\epsilon \:(x^m-x^n+x^n) \\ &=\epsilon \:x^m \\ &\leqslant \epsilon \end{align} So by Cauchy Criterion, $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}x^k$ converges uniformly on $[0,1]$.