# Determine whether following series converges:: My thoughts: When n approaches infinity so given series diverges by Comparison test because $\frac{1}{n}$ diverges. Firstly, I wonder if I am correct on this one. And would there be another, better possible solution for this series?

• Does the series in question have $n^3$ on the denominator of the summand (as it is in the second expression) or not (as it is in the first)? It's kinda important :)
– Hugh
Apr 25, 2014 at 8:44

Your idea is basically correct and a good approach to dealing with series at this level (viz, comparison test with the harmonic series). I would say you need more intermediate steps though.

$$s_n = \frac{n^2 + 3}{n^3(2 + \sin(n\pi/2))}$$

We want to establish $s_n \geq 1/n$ so we want a lower bound on $s_n$ so we want an upper bound on the denominator of $s_n$. Now we know $\sin \leq 1$ Hence,

\begin{align*} (2 + \sin(n\pi/2)) \leq 3 \end{align*} and thus $$s_n \geq \frac{n^2 + 3}{3n^3} \geq \frac{1}{n} \quad \text{for n sufficiently large}$$

• Note of course that "$\text{for$n$sufficiently large}$ is a bit cheeky as only $n=1$ fails!
– Hugh
Apr 25, 2014 at 8:59

Note that, best case scenario, $\sin(x) = 1$, in which case the denominator is maximized. Hence, we have:

$$\frac{n^2 + 3}{n^3(2+\sin(n\pi/2))} \geq \frac{n^2 + 3}{3n^3} = \frac{1}{3n} + \frac{1}{n^3}$$

From here, a quick glance at the integral test will confirm that this series is indeed divergent since $\ln(x)$ is unbounded.

• Since the post changed, modify your good answer. Apr 25, 2014 at 8:54

You can do quite simple : just look at the denominator. There, $sin(\ldots)$ is between $-1$ and $1$ so :

$0 \leq (2 + sin(\frac{n \pi}{2})) \leq 3$ hence $\frac{n^2 + 3}{(2 + sin(\frac{n \pi}{2}))} \geq \frac{n^2 + 3}{3}$ and the term not only doesn't even go to zero, it goes to infinity and the serie surely diverges