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My prof mentioned during a proof the following statement is from Calculus but I can't seem to prove it. Any suggestions?
$$\sum\limits_{k=1}^{\infty} {\sqrt\frac{1}{\pi k}} = \sqrt\frac{1}{\pi}\sum\limits_{k=1}^{\infty} {\sqrt\frac{1}{k}} = \infty.$$
I mainly don't understand how $\sum_{k=1}^{\infty} \frac{1}{\sqrt k} = \infty$, because $\lim_{k\to \infty} \frac{1}{\sqrt k} = 0 $, so intuitively speaking the sum of the fraction should approach a set number, not $\infty$ ?
Using the comparison test, we can show that $\sum_{k = 1} ^\infty \frac{1}{\sqrt{k}}$ diverges as $\sum_{k = 1} ^\infty \frac{1}{k}$ diverges. Note that $\frac{1}{k} \leq \frac{1}{\sqrt{k}}$ for all $k \geq 1$.
Are you aware of the Condensation test? If not, you can check it out here. Based on it, since $a_{k} > 0$ and nonincreasing, the proposed series converges (diverges) iff the following series converges (diverges): \begin{align*} \sum_{k=1}^{\infty}2^{k}a_{2^{k}} = \sum_{k=1}^{\infty}2^{k}\frac{1}{\sqrt{2^{k}}} = \sum_{k=1}^{\infty}2^{k/2} = \sum_{k=1}^{\infty}(\sqrt{2})^{k} \end{align*}
Once the last series diverges, the original series also diverges.
That is because a geometric series diverges when its ratio is greater than one.
Hopefully this helps !
The first step is essentially that ${\sqrt\frac{1}{\pi k}}={\sqrt\frac{1}{\pi }}{\sqrt\frac{1}{ k}}$ since $\frac 1 \pi$ and $\frac1k$ are positive, and then taking the constant $\sqrt\frac{1}{\pi }$ outside the summation. It is not particularly interesting here.
There are several different ways to show $\sum\limits_{k=1}^{\infty} {\sqrt\frac{1}{k}}$ is infinite. One is to say $\sqrt\frac{1}{x}$ is a decreasing function of $x$ with integral $2\sqrt{x}+c$ so $ \sqrt\frac{1}{k} \ge \int\limits_{x=k}^{k+1} \sqrt\frac{1}{x} \,dx $ and thus $$ \sum\limits_{k=1}^{n} {\sqrt\frac{1}{k}} \ge \int\limits_{x=1}^{n+1} \sqrt\frac{1}{x} \,dx =2\sqrt{n+1}-2$$ and, since the right hand side increases towards infinity as $n\to \infty$, the left hand side which is larger must too. You can use a similar argument to say the partial sum is no more than $2\sqrt{n}-1$, so you now also know how fast it increases towards infinity.
Since $1/\sqrt{k}$ is a decreasing function, we can say that $$\sum_{k=1}^{N}\frac{1}{\sqrt{k}} \ge \int_{1}^{N+1}\frac{dk}{\sqrt{k}}=2\sqrt{k}\Big\vert_{1}^{N+1}=2\sqrt{N+1}-2.$$ This diverges to $\infty$ as $N\rightarrow\infty$, and so the original series also diverges.
