2^(n+1)/2(n+2)^2+1
When I found lim a(n+1)/a(n) = lim 2 (n is approaching infinity) =2
2 is greater than 1, series converges.
I probably made a mistake in lim calculations, if 2 is incorrect, will you please show me how to do it.
Thank you very much

2. Is this $\sum_{n}^\infty\frac{2^{n+1}}{2(n+2)^2+1}$?

$\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=\lim_{n\to\infty}\frac{2^{n+1}}{2(n+2)^2+1}\fr ac{2(n+1)^2+1}{2^n}=2\lim_{n\to\infty}\frac{2n^2+4 n+3}{2n^2+8n+5}=2\cdot 1 = 2.$

So the serie diverges.

In fact $a_n\geq \frac{2^{n+1}}{2(n+2)^2+2}=\underbrace{\frac{2^n}{ n^2+4n+5}}_{n\to\infty,\to\infty}$

3. ## converges or diverges

I am sorry, I typed the problem wrong, it is 2^n/2n^2 + 1. I am very sorry. Can you explain this one.
Thank you

4. It's easier than it's looks like: If the serie converges $\Rightarrow$ $a_n\to 0$ then if $\neg(a_n\to 0)\Rightarrow$ the serie diverges.

$\lim_{n\to\infty}\frac{2^n}{2n^2+1}=\lim_{n\to\inf ty}\frac{2^n\log 2}{4n}=\lim_{n\to\infty}\frac{2^n\log 2\log 2}{4}\to\infty$

5. ## Please solve it by ratio test

I am sorry, but I do not understand it.

Will you please show how to solve it by ratio test 2^n/2n^2 + 1.
I did lim a(n+1) /a(n)= lim 2 =2
The series diverges because 2 is greater than 1
Is this correct?

6. Yes, that's correct, but we can just bound the general term.

For each $n\ge1$ it's $\frac{2^{n}}{2n^{2}+1}>\frac{n}{2n^{2}+2n^{2}}=\fr ac{1}{4}\cdot \frac{1}{n},$ which diverges by direct comparison test with the harmonic series.

7. ## Ratio Test

Will you please check for correctness. Thank you a lot.

Determine whether the series is convergent or divergent by Ratio Test.

series (n+1)/n^2
By Ratio Test lim is 1 , so the ratio test failed. We can use Limit Comparison. By Limit Comparison Test lim is 1, which is greater than 0, so both series are divergent based on b(n), because b(n) is divergent.

Please explain if I am wrong. Thank you again

8. That's correct, provided that $b_n=\frac1n.$

You can also bound: $\frac{n+1}{n^{2}}> \frac{n}{n^{2}}=\frac{1}{n},$ and the conclusion follows.