Summation of an Infinite Series

[jetter2]

So I have the following two problems that I am able to take the limit of to see if they converge or diverge, my issue is for the ones that DO converge, how do I find their sum?



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#6 is an alternating series, and I am able to take the limit to find it converges to zero..my issue is HOW do I find the sum?
#7 converges to zero as well, how can I find the sum?
#8 If I apply L'hospitals rule I get to 1/2 and can tell it is divergant.


Can someone help point me in the right direction?

 

[galactus]

10 diverges.

You can find the sum of 6 and 8 by using the geometric series.

Using \(\displaystyle \sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}\),


#6: Rewrite as \(\displaystyle \frac{-5}{6}\sum_{n=1}^{\infty}\left(\frac{-6}{5}\right)^{n}\)

\(\displaystyle \frac{-5}{6}\cdot \frac{\frac{-6}{5}}{1+\frac{6}{5}}=\frac{5}{11}\)

#8: \(\displaystyle \sum_{n=0}^{\infty}\left(\frac{1}{\sqrt{2}}\right)^{n}\)

sub in \(\displaystyle x=\frac{1}{\sqrt{2}}\)

\(\displaystyle \frac{1}{1-\frac{1}{\sqrt{2}}}=\sqrt{2}+2\)


Finding the sums of convergent series can be fun and challenging.

Such as \(\displaystyle \sum_{k=0}^{\infty}\frac{n^{2}}{3^{n}}\)

One can use the general form of the geometric series and differentiate to hammer it into the correct form to find the sum.

Or the very famous:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}}{6}\), which is not geometric and takes more clever methods.

This is Euler's famous Basel problem.

 

[jetter2]

Man I'm a bad student! I even had the problem numbers wrong LOL.


Now that you mention turning into a geometric series it makes perfect sense, let me try it out and see what I get.

Thanks again!