# Partial Sums

• April 13th 2011, 04:57 AM
Partial Sums
http://math.rutgers.edu/~greenfie/we...fstuff/w4H.pdf

I know for a), I can show it converges with integral test, but how can I find a specific partial sum?

Do I just take the integral from 2 to n of 1/(x*(ln x)^2) and set it <10^-5 so that I get 5 digit accuracy, and then solve for N?

For b), I know how to show divergence, but how do I find the specific partial sum? Same idea?
• April 13th 2011, 07:15 AM
FernandoRevilla
I don't know exactly what resources you can use but for example if an alternating series $u_1-u_2+u_3-\ldots$ satisfies the hypothesis of the Leibnitz's rule test then, $|S_n-S|\leq u_{n+1}$ .
• April 13th 2011, 08:02 AM
chisigma
Quote:

http://math.rutgers.edu/~greenfie/we...fstuff/w4H.pdf

I know for a), I can show it converges with integral test, but how can I find a specific partial sum?

Do I just take the integral from 2 to n of 1/(x*(ln x)^2) and set it <10^-5 so that I get 5 digit accuracy, and then solve for N?

For b), I know how to show divergence, but how do I find the specific partial sum? Same idea?

a) the series...

$\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n\ \ln^{2} n}$ (1)

... is 'alternate sign' and the absolute value of the genral term decreases monotonically and tends to 0 with n, so that the series converges. For this type of series the truncation error is inferior, in absolute value, to the first 'discharged term' and that happens for...

$\displaystyle n\ \ln^{2} n \sim 10^{5} \implies n \sim 1800$ (2)

Kind regards

$\chi$ $\sigma$
• April 13th 2011, 08:30 AM
chisigma
Quote:

http://math.rutgers.edu/~greenfie/we...fstuff/w4H.pdf

I know for a), I can show it converges with integral test, but how can I find a specific partial sum?

Do I just take the integral from 2 to n of 1/(x*(ln x)^2) and set it <10^-5 so that I get 5 digit accuracy, and then solve for N?

For b), I know how to show divergence, but how do I find the specific partial sum? Same idea?

b) the series...

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\ \sqrt{\ln n}}$ (2)

... diverges because is...

$\displaystyle I(x)= \int_{2}^{x} \frac{dt}{t\ \sqrt{\ln t}} = 2\ \sqrt {\ln x} - 2\ \sqrt{\ln 2}$ (2)

... and I(x) tends to infinity with x. The condition I(x)= 100 is verified for $x \sim 5.5\ 10^{1085}$ (Itwasntme)...

Kind regards

$\chi$ $\sigma$
• April 13th 2011, 09:09 AM