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enter image description here My thoughts: When n approaches infinity

enter image description here

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?

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  • $\begingroup$ 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 :) $\endgroup$
    – Hugh
    Commented Apr 25, 2014 at 8:44

3 Answers 3

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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 $$ s_n \geq \frac{n^2 + 3}{3n^3} \geq \frac{1}{n} \quad \text{for $n$ sufficiently large} $$

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

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  • $\begingroup$ Since the post changed, modify your good answer. $\endgroup$ Commented Apr 25, 2014 at 8:54
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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

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