Limit of a infinite series

Question

$$\frac{2}{2} + \frac{2\cdot 5}{2\cdot 9} + \frac{2\cdot 5\cdot 10}{2\cdot 9\cdot 28} + \cdots + \frac{2\cdot 5\cdot 10 \cdots (n^2+1)}{2\cdot 9\cdot 28\cdots (n^3+1)}\tag1$$

For this series $(1)$, how would one go about applying the comparison test to check for convergence or divergence?

Apologies for my lack of knowledge, but the test has not yet been covered in any of my classes yet. Comparison, root and ratio test are all that I know how to apply. I would study the Raabe's test — Paras Joshi, Mar 01 '17 at 04:30
It comes as in conclusive as I get a value of 1 for the ratio — Paras Joshi, Mar 01 '17 at 04:35

score 3 · Accepted Answer · answered Mar 01 '17 at 04:34

3

The general term of the series, $a_n$ is given by

$$a_n=\prod_{k=1}^n\frac{(k^2+1)}{(k^3+1)}$$

Then, we see that

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2+1}{(n+1)^3+1}\to 0\implies \text{the series converges}$$

answered Mar 01 '17 at 04:34

Mark Viola

179,405

The denominator is missing in the above calculation – Paras Joshi Mar 01 '17 at 04:43
The denominator is not missing. $$\prod_{k=1}^{n+1}f_n=f_{n+1}\prod_{k=1}^{n}f_n$$Hence$$\frac{\prod_{k=1}^{n+1}f_n}{\prod_{k=1}^{n}f_n}=f_{n+1}$$ – Mark Viola Mar 01 '17 at 04:45
1

Apologies, you are correct. Complete mistake on my end. – Paras Joshi Mar 01 '17 at 04:46
Apology not necessary - no worry. Pleased to see that you see it now. – Mark Viola Mar 01 '17 at 04:51

score 0 · Answer 2 · answered Mar 01 '17 at 04:36

For $n\ge 3$ we get $$\frac{n^2+1}{n^3+1}<\frac{n^2+1}{n^3}=\frac1n+\frac1{n^3}\le\frac13+\frac1{27}=\frac4{27}<\frac12$$ Then $$\sum_{k=1}^n\prod_{j=1}^k\frac{j^2+1}{j^3+1}<1+\frac{5}{9}+\sum_{k=3}^n\frac59\left(\frac12\right)^{k-2}$$

So, by comparison test, we get that the given series converges.

Limit of a infinite series

2 Answers2