$\displaystyle

\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

$\displaystyle

\frac{k}{k + 1}

$

but I am missing something apparently

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- Apr 7th 2009, 03:54 PMTYTYseries convergence by root test
$\displaystyle

\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

$\displaystyle

\frac{k}{k + 1}

$

but I am missing something apparently - Apr 7th 2009, 04:04 PMmr fantastic
If you apply the Cauchy root test then you need to find the following limit:

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

It is not too hard to show that the value of this limit is equal to $\displaystyle e^{-1} = \frac{1}{e}$. - Apr 7th 2009, 04:17 PMTYTY
re: bolded - maybe for you lol.

thank you mister fantastic

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

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

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, 04:26 PMmr fantastic
1. $\displaystyle \left[ \left( \frac{k}{k+1}\right)^{k^2}\right]^{1/k} = \left( \frac{k}{k+1}\right)^{\frac{k^2}{k}} = \left( \frac{k}{k+1}\right)^{k}$.

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

3. Use the well known limit from calculus: $\displaystyle \lim_{n \rightarrow + \infty} \left(1 + \frac{a}{n} \right)^n = e^a$. - Apr 7th 2009, 04:33 PMTYTY
You're the man! :)

many many thanks