# series convergence by root test

• Apr 7th 2009, 04:54 PM
TYTY
series convergence by root test
$
\left(\frac{k}{k + 1}\right)^{k^2}
$

The book answer is confusing to me - I would think you could just take the limit of

$
\frac{k}{k + 1}
$

but I am missing something apparently
• Apr 7th 2009, 05:04 PM
mr fantastic
Quote:

Originally Posted by TYTY
$
\left(\frac{k}{k + 1}\right)^{k^2}
$

The book answer is confusing to me - I would think you could just take the limit of

$
\frac{k}{k + 1}
$

but I am missing something apparently

If you apply the Cauchy root test then you need to find the following limit:

$\lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k$.

It is not too hard to show that the value of this limit is equal to $e^{-1} = \frac{1}{e}$.
• Apr 7th 2009, 05:17 PM
TYTY
Quote:

Originally Posted by mr fantastic
If you apply the Cauchy root test then you need to find the following limit:

$\lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k$.

It is not too hard to show that the value of this limit is equal to $e^{-1} = \frac{1}{e}$.

re: bolded - maybe for you lol.

thank you mister fantastic

I will wrestle with it and also try to figure out why you need

$\lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k$.

This problem definitely touches on some of my weaker areas. Also thanks for the latex fix, now I learned something new there too
• Apr 7th 2009, 05:26 PM
mr fantastic
Quote:

Originally Posted by TYTY
re: bolded - maybe for you lol.

thank you mister fantastic

I will wrestle with it and also why you need

$\lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k$.

This problem definitely touches on some of my weaker areas. Also thanks for the latex fix, now I learned something new there too

1. $\left[ \left( \frac{k}{k+1}\right)^{k^2}\right]^{1/k} = \left( \frac{k}{k+1}\right)^{\frac{k^2}{k}} = \left( \frac{k}{k+1}\right)^{k}$.

2. $\frac{k}{k+1} = \frac{(k + 1) - 1}{k+1} = \frac{k+1}{k+1} - \frac{1}{k+1} = 1 - \frac{1}{k+1}$.

3. Use the well known limit from calculus: $\lim_{n \rightarrow + \infty} \left(1 + \frac{a}{n} \right)^n = e^a$.
• Apr 7th 2009, 05:33 PM
TYTY
You're the man! :)

many many thanks