If sequence $a_n$ is converging to $0$ by monotonically decreasing then $\sum_{n=1}^{\infty}a_n$ and $\sum_{k=1}^{\infty}2^ka_{2^k}$ are converging same time.
From lecture notes could understand proof of this theorem.Question is problem regarding to this theorem.
Prove that $\sum_{n=2}^{\infty}\frac{1}{n\ln{n}}$ is not converging.
proof in my book is.
$2^ka_{2^k}=2^k\frac{1}{2^k\ln{2^k}}=\frac{1}{\ln{2}}$$\frac{1}{k}$ from here how it follows that is not converging?
This is an application of the Cauchy condensation test: if $(a_n)_{n\in\Bbb N}$ is decreasing and each $a_n$ is non-negative, then $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=0}^\infty 2^na_{2^n}$ converges. In your case, the series $\sum_{n=1}^\infty2^n\frac1{2^n\log(2^n)}$ diverges, and therefore so does the series $\sum_{n=2}^\infty\frac1{n\log(n)}$.
A second mechanism for determining convergence is the integral test.
If $f : [k, \infty) \to \mathbb{R}$ is monotonically decreasing and nonnegative, then $\sum\limits_{n = k}^\infty f(n)$ converges if and only if $\int\limits_k^\infty f(x) dx$ converges, and the sum diverges to $\infty$ if and only if the integral diverges to $\infty$.
The function $f(x) = \frac{1}{x \ln x}$ is nonnegative and decreasing on $[2, \infty)$. So let's evaluate $\int\limits_2^\infty f(x) dx$.
We have $\int f(x) dx = \int \frac{1}{x \ln x} dx$. Substituting $u = \ln x$ gives us $\int \frac{1}{u} du = \ln u + C = \ln (\ln x) + C$. So $\int\limits_2^\infty f(x) dx = \lim\limits_{x \to \infty} \ln(\ln(x)) - \ln(\ln(2)) = \infty$.
Thus, the sum diverges to infinity as well.
The integral test is nice because it's broadly applicable (unlike Cauchy condensation, which is not as widely used in practice).
