Prove that $$\lim_{n\to \infty}\left(\ln 2 -\left(-\frac12+\frac13-\frac14+\cdots -\frac{(-1)^n}n\right)\right)^n =\sqrt{e}$$
I happened to encounter this problem proposed by Mohammed Bouras,Morocco in the facebook group of Romanian mathematical Magazine
As per the title, I think the limit of the problem depends upon the parity of $n$. That is,if $n$ is even, the limit is $\frac1{\sqrt e}$ otherwise as stated.
My query is, Does the parity indeed matters for this problem? And if it matters what should be the conclusion for limit of the problem?
Here is my try
we will show that the there exist two different limits for above problem.
For $0< x\leq 1$, we define the functions $$f(x)=\ln(1+x),\; \displaystyle g(x)=\sum_{k=1}^n \frac{(-x)^k}{k+1}$$ and we note that $$\begin{aligned}f(x)-g(x) &= x-\sum_{k=2}^\infty(-1)^{k+n} \frac{x^{k+n}}{k+n}\\&=x+\sum_{k=2}^{\infty} (-1)^{k+n} \int_0^x t^{k+n-1}dt\\&=x+(-1)^n\int_0^x t^n\left(\sum_{k=1 }^\infty(-1)^k t^{k-1} \right)dx\\&=x-(-1)^n\int_0^x\frac{t^n}{1+t} dt\end{aligned}$$ hence for $x=1$ we have then $$f(1)-g(1)=\ln(2)-\sum_{k=1}^\infty\frac{(-1)^k}{k+1}=1-(-1)^n\int_0^1\frac{t^n}{1+t}dt$$ Note that latter integral is know result however,here we shall derive it and we shall show that
$$\displaystyle\lim_{n\to\infty}(f(1)-g(1))^n =\begin{cases}\sqrt{e}\; \text{if } \, n\in 2n-1 \\ \frac1{\sqrt{e}} \; \text{otherwise}\end{cases}$$
We solve the following integral for any $n>0$. By polynomial long division it is trivial to note that $$\int_0^1\frac{t^n}{t+1}dt=(-1)^n\int_0^1\left(\frac{1}{t+1}-\sum_{0\leq j\leq n}(-1)^j t^{j-1}\right)dt$$ and hence on integrating $\displaystyle \int_0^1\frac{t^n}{1+t}dt$ $$\begin{aligned}&=(-1)^n\left(\log(2) -\sum_{1\leq j\leq n} \frac{(-1)^{j+1}}{j}\right)\\&=2^{-1}\left(-\psi\left(\frac{n+1}2\right)+\psi\left(\frac{2n+1}2\right)\right)\\&=\frac12\left(H_{\frac{n}2}-H_{\frac{n-1}2}\right)\end{aligned}$$ Further we note that $H_n\approx \gamma +\ln n +\frac1{2n}-O(n^{-2})$ with which we deduce that $$H_{\frac{n}2} -H_{\frac{n-1}2} \approx \frac1n-\ln\left(\frac{n-1}n\right)+\frac1{n-1}$$ for all $n>1$ and hence $H_{\frac{n}{2}} -H_{\frac{n-1}2} \to \frac1n$as $n$ gets larger. Thus we have for $$\lim_{n\to\infty}(f(1)-g(1))^n= \lim_{n\to\infty} \left(1-\frac{(-1)^n}{2n}\right)^n=e^{-\frac{(-1)^n}2} =\sqrt{e^{-(-1)^n}}$$
therefore if $n$ is even we have limit as $\displaystyle \frac1{\sqrt{e}}$ and if $n$ is odd we have limit $ \displaystyle \sqrt{e}$.
Since we have two different limits. Does it have limit ?
Thank you
Best Answer
The alternating series
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$$ is well-known to tend to $\log 2$, and the expression inside the main parenthesis oscillates around $1$. One can expect an asymptotic behavior like
$$1\pm\frac1{2n}.$$
Then taking the $n^{th}$ power, the value will alternatively rejoin $e^{1/2}$ and $e^{-1/2}$, so the limit of the sequence does not exist.
More precisely, if we group the terms in pairs, we have alternatively
$$S_{2n}=1+\sum_{k=2n+2}^\infty\frac1{2k(2k+1)}\sim 1+\frac1{4n}$$
and
$$S_{2n+1}=1+\sum_{k=2n+2}^\infty\frac1{2k(2k+1)}-\frac1{2n+1}\sim 1-\frac1{4n},$$ approximating the sums by integrals.
Taking the power, we have
$$S_{2n[+1]}^{2n}\sim\left(1\pm\frac1{4n}\right)^{2n}\sim e^{\pm1/2}.$$