Finding convergence of a series???

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May 27th 2009, 02:33 AM
Jonny-123

Finding convergence of a series???

ive run into the following question and i really don't know where to start. (Crying) i havnt come across any questions involving 2 variables yet.

For what values of x does the following series converge?

k=1(sigma)infinity = x^k/((4^k)*k)

any help would be great. (Nod)
May 27th 2009, 03:20 AM
Deadstar

What you want to do is treat x as a constant and do the ratio test.

So for $a_k = \frac{x^k}{4^k k}$ we get

$|\frac{a_{k+1}}{a_k}| = |\frac{x^{k+1}}{4^{k+1} (k+1)} \cdot \frac{4^k k}{x^k}|$ = $|\frac{k}{k+1} \cdot \frac{1}{4} \cdot x|$ .

Now for this to converge you want $|\frac{k}{k+1} \cdot \frac{1}{4} \cdot x| <1$ when k goes to infinity. So when k does go to infinity $\frac{k}{k+1} -> 1$ so we are left with $|\frac{1}{4} \cdot x| <1$
hence $-4<x<4$ .
May 27th 2009, 03:22 AM
CaptainBlack

Quote:

Originally Posted by Jonny-123

ive run into the following question and i really don't know where to start. (Crying) i havnt come across any questions involving 2 variables yet.

For what values of x does the following series converge?

k=1(sigma)infinity = x^k/((4^k)*k)

any help would be great. (Nod)

First improve your notation, more readable would be:

sum( x^k/(k 4^k) , k=1 .. infty)

better yet learn to use the math type setting facility on this site, see the tutorial here

The general term of your series is:

$a_k=\frac{(x/4)^k}{k}$

when $|x|<4$ this is bounded in absolute value by $(|x|/4)^k$ which is the general term of a convergent geometric series and hence your series is convergent.

when $x=4$ the series is the harmonic series and hence divergent, when $x=-4$ the series is the alternating harmonic series and hence convergent.

When $|x|>4$ , $a_k$ does not go to zero as $k$ goes to infinity hence the series diverges.

CB