# Prove that $\sum_{n=1}^{\infty}\frac{n^{10}}{1.1^n}$ converges

I'm trying to prove that $\displaystyle \sum_{n=1}^{\infty}\frac{n^{10}}{1.1^n}<\infty$.

My try:

Let $n_0\in \mathbb{N}$ such that for any $n>n_0$ the condition $1.1^n>n^{12}$ holds.

Now, $$\sum_{n=1}^{\infty}\frac{n^{10}}{1.1^n}=\sum_{n=1}^{n_0}\frac{n^{10}}{1.1^n}+\sum_{n=n_0+1}^{\infty}\frac{n^{10}}{1.1^n}\le\sum_{n=1}^{n_0}\frac{n^{10}}{1.1^n}+\sum_{n=n_0+1}^{\infty}\frac{n^{10}}{n^{12}}$$Both sums in RHS converge, hence $\displaystyle \sum_{n=1}^{\infty}\frac{n^{10}}{1.1^n}$ converges.

Is my reasoning fine?

By the way, According to WolframAlpha the series converges by the ratio test, but I couldn't find proper series to check with. Please explain.

Thank you!

• When using the ratio test you do not need to pit the series against another one - that business is for comparison test (aka majorant/minorant). You only need to study the limit of the ratio of two consecutive terms. Here $$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{10}}{1.1\cdot n^{10}}.$$ Oh, and your reasoning is fine. Provided that you can justify the existence of such an $n_0$. Presumably you have, at your disposal, a general result stating that exponential growth always wins against polynomial growth. – Jyrki Lahtonen Nov 17 '15 at 10:39
• @JyrkiLahtonen, thank you for useful answer. – Galc127 Nov 17 '15 at 10:43
• Another way of solving this is the root test and noticing that $n^{1/n} \rightarrow 1$ as $n$ goes off to $\infty$. – MathNewbie Nov 17 '15 at 10:44
• Actually, you have found a proper series to check with for the direct comparison test -- what you're showing here is that $a_n < n^{-2}$ for sufficiently large $n$, and you already know that $\sum n^{-2}$ converges. – Henning Makholm Nov 17 '15 at 10:56

Yes, this reasoning is fine, provided that you have a reason to know that your $n_0$ exists.
For the ratio test: The ratio between successive terms in the series is $$\frac{(n+1)^{10}/1.1^{n+1}}{n^{10}/1.1^n} = \frac{(n+1)^{10}}{n^{10}\cdot 1.1} = \frac1{1.1}\Bigl(\frac{n+1}n\Bigr)^{10} = \frac1{1.1}\Bigl(1+\frac1n\Bigr)^{10}$$ which clearly converges toward $\frac1{1.1}$ which is strictly less than $1$. Thus, the series must converge.
From the first sentence below "My Try" I infer that you know the fact that for all $a,b > 0$ we have $x^{a}/e^{bx} \to 0$ as $x \to \infty$. Using this fact we have $$\frac{n^{10}}{(1.1)^{n}} = \frac{n^{10}}{\exp [n \log (1.1)]} \leq \frac{n^{10}}{n^{12}} = n^{-2}$$ for large $n$; the series $\sum_{n \geq 1}n^{-2}$ converging implies, by comparison test, that the series under consideration converges.