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")

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- Nov 15th 2008, 03:04 AMJojo123Converge 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")

- Nov 15th 2008, 04:36 AMSoroban
Hello, Jojo123!

Quote:

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

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

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

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

. . . . . .$\displaystyle 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 $

. . . . . .$\displaystyle 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.

- Nov 15th 2008, 08:36 AMMathstud28
- Nov 15th 2008, 09:00 AMMoo
- Nov 15th 2008, 09:03 AMMathstud28