Finding the limit as n goes to infinty

**ein** · April 25th 2010, 11:22 PM

I've been trying to puzzle out how to find this limit but I am well and truly stumped. I would really appreciate the help...

Find the limit as n goes to infinity of (e^(1/n)+e^(2/n)+e^(3/n)+........+e^(n/n))/n

Thanks!

**simplependulum** · April 26th 2010, 12:31 AM

Originally Posted by ein

I've been trying to puzzle out how to find this limit but I am well and truly stumped. I would really appreciate the help...

Find the limit as n goes to infinity of (e^(1/n)+e^(2/n)+e^(3/n)+........+e^(n/n))/n

Thanks!

HI ,

If you have learnt a topic called Riemann Sum , I advice to apply it to obtain the limit .

If you not or if you feel bored about this method , i suggest solving it in this way :

$\lim_{n\to\infty} \frac{1}{n} (e^{1/n} + e^{2/n} + e^{3/n} + ... + e^{n/n} )$

$= \lim_{n\to\infty} \frac{1}{n} (t + t^2 + t^3 + ... + t^n )$ where $t = e^{1/n}$

$= \lim_{n\to\infty} \frac{1}{n} \frac{t( t^n - 1)}{t-1}$

Since $\lim_{n\to\infty} e^{1/n} ( or ~ t) = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^{n/n} =\lim_{n\to\infty} (1 + \frac{1}{n} )$

The limit is

$\lim_{n\to\infty} \frac{1}{n} (1+\frac{1}{n}) \frac{e-1}{1+\frac{1}{n} - 1 }$

$= \lim_{n\to\infty} \frac{1}{n} n(1+ \frac{1}{n})(e-1)$

$= \lim_{n\to\infty} (1+\frac{1}{n})(e-1) = e-1$

**Prove It** · April 26th 2010, 12:41 AM

Another alternative is to realise that the numerator is a geometric series.

Writing it in closed form, i.e. $\frac{a(1 - r^n)}{1 - r}$ will simplify matters greatly.

Thread: Finding the limit as n goes to infinty

LinkBack

Thread Tools

Display

Finding the limit as n goes to infinty

Similar Math Help Forum Discussions

Limits involving products of infinty and zero.

Finding limit of this function, using Limit rules

limit n-->infinty

product of n unifom distribution as n-> infinty

continuous at infinty?

Search Tags