# Thread: Help in Interval of convergence problems!

1. ## Help in Interval of convergence problems!

Q1) I need help in finding the open interval of convergence for the series..

--- [(k!) (x^k)] / (k^k) ........when k goes from 1 to infinity..Note: do not examine convergence at the endpoint of the interval.

Q2) Find all the values of the x for which the series converges!

[(x+2)^2k ] / (2^k) using the ratio test...

I solved the function using ratio test, but I did not know how to proceed after then...

2. Originally Posted by Vedicmaths
Q1) I need help in finding the open interval of convergence for the series..

--- [(k!) (x^k)] / (k^k) ........when k goes from 1 to infinity..Note: do not examine convergence at the endpoint of the interval.
[snip]
Using the ratio test you should be able to get

$|x| < \frac{1}{L}$ where $L = \lim_{k \rightarrow \infty} \frac{k^k}{(k+1)^k}$.

One way of finding this limit is to note that

$\frac{k^k}{(k+1)^k} = \left( \frac{k}{k+1} \right)^k = \left(1 - \frac{1}{k+1} \right)^k = \left(1 + \frac{(-1)}{k+1} \right)^k$.

The limit is now seen to be a standard form ......

3. Originally Posted by Vedicmaths
[snip]
Q2) Find all the values of the x for which the series converges!

[(x+2)^2k ] / (2^k) using the ratio test...

I solved the function using ratio test, but I did not know how to proceed after then...
Can you get to $\lim_{k \rightarrow \infty} \left| \frac{(x+2)^{k+1}}{2}\right| < 1$?

So you need $|x+2| < 1$ otherwise the limit 'blows up'. Now you need to check what happens when $|x+2| = 1$, that is, when x = -1 and x = -3 .....

4. I got (x+1)^2 / 2....but I guess it won't change the meaning of what you were explaining. so does that mean for the value of x = -1 and -3 the series converges because of the fact that we need to show that |x+2| < 1?

5. Originally Posted by Vedicmaths
I got (x+1)^2 / 2....but I guess it won't change the meaning of what you were explaining. so does that mean for the value of x = -1 and -3 the series converges because of the fact that we need to show that |x+2| < 1?
Actually the series will converge when |x + 2| = 1 because $\lim_{k \rightarrow \infty} \left| \frac{(\pm1)^{k+1}}{2}\right| = \frac{1}{2} < 1$.

6. ## First one!

thanks for the help for the second part..

In the first, I am getting (x) (k^k) / (k+1)^k...will that change everything??
And the one you showed below, I think that will lead us to the exponent form e^x right??

thanks for the help Sir!

7. Originally Posted by Vedicmaths
thanks for the help for the second part..

In the first, I am getting (x) (k^k) / (k+1)^k...will that change everything??
And the one you showed below, I think that will lead us to the exponent form e^x right??

thanks for the help Sir!
$\frac{\frac{(k+1)! \, x^{k+1}}{(k+1)^{k+1}}}{\frac{k! \, x^k}{k^k}} = \frac{(k+1)! \, x^{k+1} \, k^k}{k! \, x^k \, (k+1)^{k+1}} = \frac{(k+1) \, x \, k^k}{(k+1)^{k+1}} = \frac{x \, k^k}{(k+1)^k}$.

Using the ratio test you therefore require

$\lim_{k \rightarrow \infty} \left| \frac{x \, k^k}{(k+1)^k} \right| < 1$ $\Rightarrow |x| < \frac{1}{\lim_{k \rightarrow \infty} \frac{k^k}{(k+1)^k}}$

which is essentially a less pleasant looking version of what I gave in post #2.

The value of the limit is $e^{-1} = \frac{1}{e}$.

8. wow!

That was awesome!

Thanks a lot!