# Thread: limit of sequence

1. ## 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?

2. ## Re: limit of sequence

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 ?

I think I misread your question a little.

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

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

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

It is of course a "series", rather than a sequence.

3. ## Re: limit of sequence

Originally Posted by Archie Meade
Do you understand how the first term and common ratio of the geometric series,
were placed into the formula for summing the series ?
Oh goodness, I am a moron. Thanks, I see what happened. I guess they assumed that would have been at the surface of my mind from when I was in high school. not!

4. ## Re: limit of sequence

It may have been offputting to see "n" as the term in the numerator and denominator
as it is also the number of terms in the series.