W WartonMorton Jan 2010 72 2 May 13, 2010 #1 Does this series converge or diverge? [(2k+1)^(2k)]/[(5k^2+1)^k]

S Sudharaka Dec 2009 872 381 1111 May 13, 2010 #2 WartonMorton said: Does this series converge or diverge? [(2k+1)^(2k)]/[(5k^2+1)^k] Click to expand... Dear WartonMorton, \(\displaystyle \frac{(2k+1)^{2k}}{(5k^2+1)^k}=\left(\frac{(2k+1)^2}{5k^2+1}\right)^k\) Now divide \(\displaystyle (2k+1)^2~and~5k^2+1~by~k^2\) Hope you can continue from here. Reactions: WartonMorton

WartonMorton said: Does this series converge or diverge? [(2k+1)^(2k)]/[(5k^2+1)^k] Click to expand... Dear WartonMorton, \(\displaystyle \frac{(2k+1)^{2k}}{(5k^2+1)^k}=\left(\frac{(2k+1)^2}{5k^2+1}\right)^k\) Now divide \(\displaystyle (2k+1)^2~and~5k^2+1~by~k^2\) Hope you can continue from here.

W WartonMorton Jan 2010 72 2 May 13, 2010 #3 Sudharaka said: Dear WartonMorton, \(\displaystyle \frac{(2k+1)^{2k}}{(5k^2+1)^k}=\left(\frac{(2k+1)^2}{5k^2+1}\right)^k\) Now divide \(\displaystyle (2k+1)^2~and~5k^2+1~by~k^2\) Hope you can continue from here. Click to expand... Once I do that I am seeing convergence?

Sudharaka said: Dear WartonMorton, \(\displaystyle \frac{(2k+1)^{2k}}{(5k^2+1)^k}=\left(\frac{(2k+1)^2}{5k^2+1}\right)^k\) Now divide \(\displaystyle (2k+1)^2~and~5k^2+1~by~k^2\) Hope you can continue from here. Click to expand... Once I do that I am seeing convergence?

S Sudharaka Dec 2009 872 381 1111 May 13, 2010 #4 WartonMorton said: Once I do that I am seeing convergence? Click to expand... Dear WartonMorton, Is it so? The term inside the bracket converges to 2/5 (when k tends to infinity). But the whole term has a \(\displaystyle k^{th}\) power. So.....

WartonMorton said: Once I do that I am seeing convergence? Click to expand... Dear WartonMorton, Is it so? The term inside the bracket converges to 2/5 (when k tends to infinity). But the whole term has a \(\displaystyle k^{th}\) power. So.....