Convergence of power series

[Melissa00]

Hi
How can I show that this power series converges? I tried to solve they problem by writing eˆn out as a series as well, but it made everything more complicated.

∑ 1/(n*[FONT=monospace, Courier]eˆn) with n=0 and k=infinity[/FONT]

Will looking at it separately help me? Obviously 1/n diverges and 1/(e[FONT=monospace, Courier]ˆn) converges, right?
I'd greatly appreciate any help [/FONT][FONT=monospace, Courier]
[/FONT]

 

[mathmari]

Let $\displaystyle a_{n}=\frac{1}{ne^n}$
Take absolute values and apply the Ratio Test:
$\displaystyle \lim_{n\rightarrow \infty } \frac{a_{n+1}}{a_{n}}$ = $\displaystyle \lim_{n\rightarrow \infty } \frac{\frac{1}{(n+1)e^{n+1}}}{\frac{1}{ne^n}}$ = $\displaystyle \lim_{n\rightarrow \infty } \frac{ne^n}{(n+1)e^ne}$ = $\displaystyle \lim_{n\rightarrow \infty } \frac{n}{(n+1)e}$ =$\displaystyle \lim_{n\rightarrow \infty } \frac{1}{(1+\frac{1}{n})e}=\frac{1}{e}<1$
So the power series converges!

 

[Melissa00]

Thanks

 

[lookagain]

Hi
How can I show that this power series converges? I tried to solve they problem by writing eˆn out as a series as well,

but it made everything more complicated.
> > > ∑ 1/(n*eˆn) with n=0 and k=infinity < < <

Will looking at it separately help me? Obviously (?) the summation from n = 1 to 00 of 1/n diverges and

the summation from n = 1 to oo of 1/(eˆn) converges, right?

Right. $\displaystyle \ \ \$ ("Obviously" is a relative term.)


$\displaystyle \displaystyle\sum_{n = 0}^{oo} \ \dfrac{1}{ne^n} \ \$


diverges, just from the first term alone, because when n = 0, the first term is undefined.

Did you mean to sum from n = 1 to oo?

 

[lookagain]

$\displaystyle \displaystyle\sum_{n = 1}^{oo} \ \dfrac{1}{ne^n} \ \$ *does* converge.


In addition to the method that mathmari used, let's look at the ...

---------------------------------------------------------------

p-Series Convergence test

The p-series is given by
Summation from n = 1 to oo of (1/n^p) = 1/1^p + 1/2^p + 1/3^p + ...
where p > 0 by definition.
If p > 1, then the series converges.
If 0 < p <= 1 then the series diverges.


Source: http://www.math.com/tables/expansion/tests.htm



----------------------------------------------------------------


Look at $\displaystyle ne^n,$ which is the denominator of the integrand.

Show that for all positive integers n, $\displaystyle \ ne^n \ = \ n^{(1 + a)}, \ \ \ where \ \ a > 0. \ \ \ \$(a varies as n varies.)

If this can be shown, then the amended series, with the limits of summation from n = 1 to oo, converges by the
p-Series Convergence test.

Knowing here that n takes on all of the values of the positive integers, it is relatively common knowledge that $\displaystyle e^n > n, \ \ or$

that fact can be established by mathematical induction, as a for instance.

Then $\displaystyle ne^n > n(n) = n^2.$

Because p, the exponent for the base n, is at least greater than 1, then the amended series converges.

​

 

[Melissa00]

Thank you so much