# limit of sequence

• Jun 28th 2011, 04:38 PM
zg12
limit of sequence
I need to show that the limit of the sequence,

$\displaystyle \frac{{n + 1}}{{{n^2}}} + \frac{{{{(n + 1)}^2}}}{{{n^3}}} + \cdots + \frac{{{{(n + 1)}^n}}}{{{n^{n + 1}}}}$

is e-1. What's sad is I have the solution in front of me and I can't understand the simple algebra.

$\displaystyle \frac{{n + 1}}{{{n^2}}}\left[ {\frac{{1 - {{\left( {\frac{{n + 1}}{n}} \right)}^n}}}{{1 - \frac{{n + 1}}{n}}}} \right] = \frac{{n + 1}}{n}\left[ {{{\left( {1 + \frac{1}{n}} \right)}^n} - 1} \right]$

The second factor on the left, how did they get that?
• Jun 28th 2011, 04:49 PM
Re: limit of sequence
Quote:

Originally Posted by zg12
I need to show that the limit of the sequence,

$\displaystyle \frac{{n + 1}}{{{n^2}}} + \frac{{{{(n + 1)}^2}}}{{{n^3}}} + \cdots + \frac{{{{(n + 1)}^n}}}{{{n^{n + 1}}}}$

is e-1. What's sad is I have the solution in front of me and I can't understand the simple algebra.

$\displaystyle \frac{{n + 1}}{{{n^2}}}\left[ {\frac{{1 - {{\left( {\frac{{n + 1}}{n}} \right)}^n}}}{{1 - \frac{{n + 1}}{n}}}} \right] = \frac{{n + 1}}{n}\left[ {{{\left( {1 + \frac{1}{n}} \right)}^n} - 1} \right]$

The second factor on the left, how did they get that?

One "n" in the denominator of the left factor was taken inside the brackets.

$\displaystyle \frac{{n + 1}}{{{n^2}}}\left[ {\frac{{1 - {{\left( {\frac{{n + 1}}{n}} \right)}^n}}}{{1 - \frac{{n + 1}}{n}}}} \right] = \frac{n+1}{n}\left[\frac{1-\left(\frac{n+1}{n}\right)^n}{n\left(1-\frac{n+1}{n}\right)}\right]$

$\displaystyle = \frac{n+1}{n}\left[\frac{1-\left(\frac{n+1}{n}\right)^n}{n\left(\frac{n-n-1}{n}\right)}\right]$

and the rest follows.

Do you understand how the first term and common ratio of the geometric series,
were placed into the formula for summing the series ?

$\displaystyle a=u_1=\frac{n+1}{n^2}$

$\displaystyle r=\frac{n+1}{n}$

$\displaystyle S_n=\frac{a\left(1-r^n\right)}{1-r}$

It is of course a "series", rather than a sequence.
• Jun 28th 2011, 05:04 PM
zg12
Re: limit of sequence
Quote: