Root Test for Series: sum [k=1 to k=infty] [(k/k+1)^(2k^2)]

[ohayme]

Use Root Test to determine whether the following series converge.

Sigma( index k=1 to k= infinity) of [(k/k+1)^(2k^2)]

Please help!! I only know that after taking the root the series become [(k/k+1)^2k] but I dont know how to take the limit of that.

The answer says the limit is e^-2, but I dont know how they got there.

 

[pka]

Use Root Test to determine whether the following series converge.
Sigma( index k=1 to k= infinity) of [(k/k+1)^(2k^2)]
Please help!! I only know that after taking the root the series become [(k/k+1)^2k] but I dont know how to take the limit of that. The answer says the limit is e^-2, but I dont know how they got there.

$\displaystyle {\lim _{k \to \infty }}\sqrt[k]{{{{\left( {\frac{k}{{k + 1}}} \right)}^{2{k^2}}}}} = {\lim _{k \to \infty }}{\left( {\frac{k}{{k + 1}}} \right)^{2k}} = {\lim _{k \to \infty }}{\left( {{{\left( {1 - \frac{1}{{k + 1}}} \right)}^k}} \right)^2} = {e^{ - 2}}$

 

[ohayme]

$\displaystyle {\lim _{k \to \infty }}\sqrt[k]{{{{\left( {\frac{k}{{k + 1}}} \right)}^{2{k^2}}}}} = {\lim _{k \to \infty }}{\left( {\frac{k}{{k + 1}}} \right)^{2k}} = {\lim _{k \to \infty }}{\left( {{{\left( {1 - \frac{1}{{k + 1}}} \right)}^k}} \right)^2} = {e^{ - 2}}$

Okay. I see now. Thanks!!