# infinite series Euler's constant

• Feb 18th 2011, 11:53 AM
mathsohard
infinite series Euler's constant
(a) If Xn=1-1/2+1/3-...-1/(2n), show that Xn -> log 2 as n -> oo, by using the definition of Euler's constant. Hint: Show that Xn=C2n-Cn + log2 in the notation

(b) Prove that the partial sums of the series

1+1/2-1/3+1/4+1/5-1/6++- ...

approach + oo as n -> oo. Make use of Euler's constant

Help
• Feb 18th 2011, 12:20 PM
chisigma
Quote:

Originally Posted by mathsohard
(a) If Xn=1-1/2+1/3-...-1/(2n), show that Xn -> log 2 as n -> oo, by using the definition of Euler's constant. Hint: Show that Xn=C2n-Cn + log2 in the notation

Using the notation...

$\displaystyle \displaystyle S_{n}= 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$

$\displaystyle \displaystyle S_{2 n}= 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n}$

$\displaystyle \displaystyle X_{2n}= 1 - \frac{1}{2} + \frac{1}{3} + ... - \frac{1}{2n}$ (1)

... it is easy to verify that is...

$\displaystyle X_{2n}= S_{2n} - S_{n}$ (2)

But by definition of Euler's constant is...

$\displaystyle \displaystyle \gamma= \lim_{n \rightarrow \infty} S_{2n} - \ln 2n = \lim_{n \rightarrow \infty} S_{2n} - \ln n - \ln 2 = \lim_{n \rightarrow \infty} S_{n} - \ln n$ (3)

... so that combining (2) and (3) we obtain...

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} X_{2n} =\lim_{n \rightarrow \infty} S_{2n} - S_{n} = \ln 2$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 18th 2011, 12:36 PM
mathsohard
need some help on (b) as well :)
• Feb 18th 2011, 01:58 PM
zzzoak
b)
It seems to me that

$\displaystyle X_{3n}=S_{3n}-\frac{2}{3}S_n$
• Feb 19th 2011, 05:25 AM
chisigma
Perfect!!!... so that is...

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} X_{3n} = \lim_{n \rightarrow \infty} (\gamma + \ln 3 + \ln n - \frac{2}{3}\ \gamma - \frac{2}{3}\ \ln n) = \lim_{n \rightarrow \infty} \frac{\gamma}{3} + \ln 3 + \frac{\ln n}{3}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$