Finding whether $\sum \frac{3n^2+2n}{2^n}$ converges or diverges

Question

I tried to find a $b_n$ for which $a_n \le b_n$ so I did:

$$\frac{3n^2+2n}{2^n} \le \frac{1}{2^n} \Leftrightarrow \\ 3n^2+2n-1 \le 0 \Leftrightarrow \\ (n+1)(3n-1) \le 0 \Leftrightarrow \\ n \le -1 \lor n \le 1/3$$

Since $b_n = (\frac{1}{2})^n$ converges, so does $a_n$ but only for $\{ n: -1 \le n \le 1/3, n \in\mathbb{N}\}$.

Is this correct? I wrote the set of values for n like that because what I understand is that n has to be a natural number that starts at 1 (or whatever number is given which in this case was one, I don't know how to write with the sum symbol). Still the answer doesn't make a whole lot of sense to me. If n has to be a natural number doesn't that set become empty?

The sum is over all $n$, so bounding the terms for a finite set of $n$ does not say anything about the convergence of the series. Try the ratio test. — robjohn, Feb 16 '20 at 14:56
https://math.stackexchange.com/questions/3463896/finding-sum-of-series-sum-infty-k-0-frack1k22k/3463901#3463901 — lab bhattacharjee, Feb 16 '20 at 14:56
@robjohn I don't understand how a series becomes convergent if it only converges for a certain set of values? — Segmentation fault, Feb 16 '20 at 14:57
You've shown that the terms of the series are less than $\frac1{2^n}$ for a limited set of $n$, that doesn't say anything about the convergence of the entire series. A different test is needed. — robjohn, Feb 16 '20 at 14:59
@robjohn I thought that when you had $a_n \le b_n$ they either both converge or diverge. — Segmentation fault, Feb 16 '20 at 14:59
If everything is positive, you have that $a_n\le b_n$, and $\sum b_n$ converges, then $\sum a_n$ converges. The statement you give is not true. — robjohn, Feb 16 '20 at 15:02
@SilenceOnTheWire: if some of the terms are negative, the condition that $a_n\le b_n$ will not imply convergence of $\sum a_n$ from the convergence of $\sum b_n$. — robjohn, Feb 16 '20 at 15:06

Rhys Hughes · Accepted Answer · 2020-03-14T08:34:53.853

Use the ratio test: $$\frac{\bigg[\frac{3n^2+8n+5}{2^{n+1}}\bigg]}{\bigg[\frac{3n^2+2n}{2^n}\bigg]}=\frac{3n^2+8n+5}{6n^2+4n}\to\frac12$$

Some clarity:

$2^n$ grows exponentially while $3n^2+2n$ grows quadratically (much slower). So, I can tell at a glance the terms will converge, hence I pick the ratio test to illustrate this point

The last part of the statement follows because of what I refer to as "Domination Leads To Irrelevancy", that is: $$\lim_{x\to\infty}\bigg[\frac{\sum_{k=0}^m a_k x^k}{\sum_{k=0}^n b_k x^k}\bigg]=\lim_{x\to\infty}\bigg[\frac{a_m}{b_n}x^{m-n}\bigg]=\begin{cases} \frac{a_mb_n}{|a_mb_n|}\infty & m>n \\ \frac{a_m}{b_n} & m=n \\ 0 & m<n \end{cases}$$

Peter Szilas · Answer 2 · 2020-02-16T16:31:47.917

1

$a_n:=\dfrac{3n^2+2n}{2^n} <\dfrac{3n^2+2n^2}{2^n}=$

$\dfrac{5n^2}{2^n}=:b_n.$

Root test for $b_n$:

$\lim_{n \rightarrow \infty}\sqrt[n]{\dfrac{5n^2}{2^n}}=1/2<1$, hence convergent.

By comparison test $\sum a_n$ connnverges.

Used :

$\lim_{n \rightarrow \infty} 5^{(1/n)}=1$;

$\lim_{n \rightarrow \infty}(n^{(1/n)})^2=1$.

edited Feb 16 '20 at 16:31

answered Feb 16 '20 at 16:06

Peter Szilas

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score 0 · Answer 3 · answered Feb 16 '20 at 15:02

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Since the exponential function grows faster than the all power function, we can write that $3n^2+2n\leq 2^{\frac{n}{2}}$ for large enough $n$. Thus by the comparsion test series converges.

answered Feb 16 '20 at 15:02

thing

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Finding whether $\sum \frac{3n^2+2n}{2^n}$ converges or diverges

3 Answers3