Integral test for series problem

**mollymcf2009** · March 24th 2009, 08:13 PM

I am not sure how I need to solve this. Can someone get me started?

How many terms of the following series would you need to add to find its sum to within 0.01?

$\sum^{\infty}_{n=2} \frac{4}{n(ln(n))^9}$

a) $n > e^{(25)^{\frac{1}{4}}}$

b) $n > e^{(50)^{\frac{1}{8}}}$

c) $n > e^{(100)^{\frac{1}{8}}}$

d) $n > e^{(25)^{\frac{1}{8}}}$

e) $n > e^{(100)^{\frac{1}{4}}}$

**matheagle** · March 24th 2009, 08:28 PM

integral test, after you let $u=\ln x$ you will get a p-integral.
You can prove that $\sum {1\over n(\ln n)^p}$ converges whenever p exceeds one.
Now we need to be careul where we start this series, that's why I left out the n=2....

**mollymcf2009** · March 24th 2009, 08:30 PM

Originally Posted by matheagle

integral test, after you let $u=\ln x$

I know, but will that give me the answer? I just wasn't sure how to test it for a specific value.

**matheagle** · March 24th 2009, 08:55 PM

I only proved it converged.
To estimate the error, well I guess you should split the sum into two parts.
One from 2 to N and the other from N+1 to infinity.
You can then bound that second sum (which is the remainder via the integral)
$\sum_{n=2}^{\infty} {4\over n(\ln n)^9}=\sum_{n=2}^N {4\over n(\ln n)^9}+<br /> \sum_{n=N+1}^{\infty} {4\over n(\ln n)^9}$ .

Call this

$\sum_{n=2}^{\infty} {4\over n(\ln n)^9}=S_N+R_N$

You want a bound on $R_N=\sum_{n=N+1}^{\infty} {4\over n(\ln n)^9}$ .

And you can play the same game with the boxes and use the integral

$\int_N^{\infty} {4dx\over x(\ln x)^9}={4\over 8(\ln N)^8}<.01$ , solve for N, giving me (b).

The you can sum up to that point.

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