Check: Convergence of an infinite series

Question

More a check than a question - I just need to ensure that my logic is correct (I always had trouble with this):

Show whether the series

$$\sum_{n=110}^{\infty}\frac{1}{3^{n}n^{3}}$$

Is divergent, convergent or absolutely convergent.

Solution - We consider:

$$\sum_{n=110}^{\infty}\frac{1}{3^{n}n^{3}} < \sum_{n=110}^{\infty}\frac{1}{n^{3}n^{3}} < \sum_{n=110}^{\infty}\frac{1}{n^{6}}$$

Hence absolutely converges since $\frac{1}{n^{6}}$ is a generalized harmonic series with $p=6$. I can do this, since $\frac{1}{n^{3}}>\frac{1}{3^{n}}$ , and if I can show a greater series converges, then the lesser series must also converge by the comparison test. This whole changing inequalities (which always seemed rather arbitrary to me) in order to show properties is the bit that I always have to get checked because I struggle with it.

Thanks all!

P.S. How do I get my mathematical expressions to appear bigger? My indices always end up squashed to oblivion.

Answer 1

Your proof is fine.

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A nice generalization of what you've done is the following:

If $\{a_n\}$ is a sequence for which $\sum_{n = N}^{\infty} |a_n|$ converges, and $\{b_n\}$ is any bounded sequence, then $$\sum_{n = N}^{\infty} |a_n b_n|$$ converges.

To prove this, let $M$ be any bound on $\{b_n\}$ and use a comparison test with the sequence $M |a_n|$. In particular, since $1/3^n$ is bounded (by $1/3^{110}$, for example), and $1/n^3$ gives rise to a convergent $p$-series, the product must also converge.

Answer 2

For one thing, the sums of both $\frac1{3^n}$ and $\frac1{n^3}$ converge, so the sum of their product certainly converges.

For a larger fraction, use dfrac instead of frac.

Here is the above fractions using dfrac:

$\dfrac1{3^n}$ and $\dfrac1{n^3}$