Limit of a Sequence

**coobe** · June 23rd 2009, 07:16 AM

apparently every computer program gives me an error finding the limit of this:

$[\frac{4n+1}{n^2}+(\frac{1}{5})^n * \frac{n}{2n+1} + (1+\frac{10}{n})^{3n}$

my result is $e^{1000}$ . looks way too strange for me

**chisigma** · June 23rd 2009, 07:39 AM

The sequence is the sum of three sequences...

a) $\frac {1+4n}{n^2}$ for which is $\lim_{n\rightarrow \infty} =0$

b) $(\frac {1}{5})^{n}\cdot \frac{n}{1+2n}$ for which is $\lim_{n\rightarrow \infty} =0$

c) $(1+\frac {10}{n})^{3n}$ for which is $\lim_{n\rightarrow \infty} (1+\frac {30}{3n})^{3n}= e^{30}$

... so that the whole limit is $e^{30}$ ...

Kind regards

$\chi$ $\sigma$

**coobe** · June 23rd 2009, 07:49 AM

hm that c part is a bit unclear for me

i thought i could write $\lim(1+\frac{10}{n})^{3n} = \lim (1+\frac{10}{n})^n*\lim (1+\frac{10}{n})^n*\lim (1+\frac{10}{n})^n$

oh damn.... i did a really embarassing mistake

basic algebra here i come again... $e^{10}*e^{10}*e^{10} = e^{30}$ and not to the power of 1000...

but thanks, needed to clarify if my thinking was right

**dax918** · June 23rd 2009, 12:16 PM

One way to solve e sequences is to use this formula:

(An)^(Bn) = (Bn)(An-1) = L -> Limit is e^L

(1 + 10/n)^(3n) = 3n(1 + 10/n -1) = 30 -> Limit is e^30.

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