Prove that $\sum_{n=1}^{\infty} \frac{(-1)^{[\sqrt n]}}{n}$ is converges where $[x]$ is the largest integer is less than or equal to x.

Question

Prove that $\sum_{n=1}^{\infty} \frac{(-1)^{[\sqrt n]}}{n}$ converges where $[x]$ is the largest integer that is less than or equal to x.

Suggestion: Use the inequality $\frac{1}{l+1}< \int_{l}^{l+1} \frac{1}{x}dx <\frac{1}{l}$.

I solved this question with Dirichlet test. https://en.wikipedia.org/wiki/Dirichlet%27s_test

Let $a_{n}=\frac{1}{n}$. It is monotonically decreasing and approaches to zero.

and $b_{n}=(-1)^{[\sqrt n]}$ and the partial sum is bounded .

By the given DT, it $ \sum_{n=1}^{\infty}a_{n}{b_{n}$ is convergent.

Can I solve this question by this approach?

Answer 1

You can apply Dirichlet's test but not to $a_n= \frac1n,b_n = (-1)^{[\sqrt n]}$ (as $\sum_{n\le x}(-1)^{[\sqrt n]}$ is not bounded), look instead at $a_n = \frac1{n^{1/3}}, b_n=\frac{(-1)^{[\sqrt n]}}{n^{2/3}}$ so that the boundedness of $\sum_{n\le x} b_n$ is obvious.

Answer 2

Note that your sum can be written as $$ \sum_{n=1}^\infty(-1)^n\sum_{k=n^2}^{n^2+2n}\frac1k\tag1 $$ Furthermore, it is simple to see that $$ \sum_{k=n^2}^{n^2+2n}\frac1k=\frac2n+O\!\left(\frac1{n^2}\right)\tag2 $$ Thus, the sum converges since the alternating harmonic series converges and $\sum\limits_{n=1}^\infty\frac1{n^2}$ converges absolutely.