# Math Help - Converge of Diverge

1. ## Converge of Diverge

Determine whether those series are converge and diverge?if possible, find the limit of the convergent series.
1. (n^+2^n)/(n+3^n) from n=1 to infinite
2. log(8n^2 + 1) + 2 log(1/n)
3. [√(n^4+ 3n^2+ 1)] - n^2 -1 ( just this sequence, not "sum")

2. Hello, Jojo123!

$2)\;\;S \;=\;\sum^{\infty}_{n=1}\bigg[\ln(8n^2 + 1) + 2\log\left(\tfrac{1}{n}\right)\bigg]$

We have: / $\ln(8n^2+1) + \ln\left(\tfrac{1}{n}\right)^2 \;=\;\ln(8n^2+1) + \ln\left(\tfrac{1}{n^2}\right) \;=\;\ln\bigg[(8n^2+1)\cdot\tfrac{1}{n^2}\bigg]$

. . . . . . $= \;\ln\left(\frac{8n^2+1}{n^2}\right) \;=\;\ln\left(\frac{8n^2}{n^2} + \frac{1}{n^2}\right) \;=\;\ln\left(8 + \frac{1}{n^2}\right)$

Hence: . $S \;=\;\sum^{\infty}_{n=1}\ln\left(8 + \tfrac{1}{n^2}\right)$

. . . . . . $S \;=\;\ln\left(8 + \tfrac{1}{1^2}\right) + \ln\left(8 + \tfrac{1}{2^2}\right) + \ln\left(8 + \tfrac{1}{3^2}\right) + \hdots$

. . . . . . $S \;=\;\ln\underbrace{\bigg[\left(8 + \tfrac{1}{1}\right)\left(8 + \tfrac{1}{4}\right)\left(8 + \tfrac{1}{9}\right) \hdots \bigg]}_{\text{infinite product}}$

This product has an infinite number of factors which are greater than 8.
. . Hence, the product diverges.

Therefore, the series diverges.

3. Originally Posted by Jojo123
Determine whether those series are converge and diverge?if possible, find the limit of the convergent series.
1. (n2+2n)/(n+3n) from n=1 to infinite
2. log(8n2 + 1) + 2 log(1/n)
3. [√(n4+ 3n2+ 1)] - n2 -1

For the other two consider if the kernel of the summation is a null sequence or not.

4. Originally Posted by Mathstud28
For the other two consider if the kernel of the summation is a null sequence or not.
What do you mean ?

5. Originally Posted by Moo
What do you mean ?
Yes, thank you very much Moo. I need to learn to say things in a more understandable fashion when helping someone online. To original poster, what I said just means that make sure that sequence being summed tends to zero. Otherwise it diverges.