series convergence by root test

**TYTY** · April 7th 2009, 03:54 PM

$ \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

**mr fantastic** · April 7th 2009, 04:04 PM

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 $e^{-1} = \frac{1}{e}$ .

**TYTY** · April 7th 2009, 04:17 PM

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

**mr fantastic** · April 7th 2009, 04:26 PM

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: $\lim_{n \rightarrow + \infty} \left(1 + \frac{a}{n} \right)^n = e^a$ .

**TYTY** · April 7th 2009, 04:33 PM

You're the man!

many many thanks

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