Examining the following Series for convergence with proof

Question

Good day dear community,

I have to examine the following series for covergence and proof it.

(i) $$\sum_{n=1}^\infty \frac{n+1}{3n}$$ (ii) $$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n}$$

(iii) $$\sum_{n=1}^\infty \frac{n^4}{3^n}$$

My approach to (i). My final step where I am stuck now: $\lim\limits_{n\to\infty}\frac{3n^2+6n}{3n^2+6n+3}$

My approach to (ii)

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

$\frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}}$

$\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0 $ because $\frac{1}{\sqrt{n}} < \frac{1}{n^2}$ So this series converges by the alternating test series.

My approach to (iii)

And here I have no idea, whatsoever.

Answer 1

For part $(i)$:

Notice that $$\sum_{n=1}^\infty\frac{n+1}{3n} \ge \sum_{n=1}^\infty\frac{n}{3n}.$$

For part $(iii)$:

Use ratio test or note that there exist $N$ such that $n \ge N,$we have $ 2^n \ge n^4$.

$$\sum_{n=1}^\infty \frac{n^4}{3^n}=\sum_{n=1}^{N-1}\frac{n^4}{3^n}+\sum_{n=N}^\infty \frac{n^4}{3^n} \le \sum_{n=1}^{N-1}\frac{n^4}{3^n}+\sum_{n=N}^\infty \left(\frac{2}{3}\right)^n$$

Answer 2

ii) grouping even-odd terms,

$$\frac1{\sqrt{2n}}-\frac1{\sqrt{2n+1}}=\frac{\sqrt{2n+1}-\sqrt{2n}}{\sqrt{2n}\sqrt{2n+1}}=\frac{1}{\sqrt{2n}\sqrt{2n+1}(\sqrt{2n+1}+\sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.

Answer 3

The (iii) converges by limit comparison test with $\sum \frac1{n^2}$ indeed

$$\frac{ \frac{n^4}{3^n}}{\frac1{n^2}}= \frac{n^6}{3^n}\to 0$$