# Alternating Series sum up to $99$ terms is $>0.5$

Prove that $$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots \cdots \cdots -\frac{1}{99}+\frac{1}{100}>0.2$$

$\bf{My\; Try::}$ We can write series as $$\frac{1}{2}\bigg(1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{50}\bigg)-\bigg(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots \cdots +\frac{1}{99}\bigg)$$

Now How can i solve after that, Help Required, Thanks

• Do you mean 0.2 or 0.5? – Henning Makholm Nov 20 '16 at 10:42
• @HenningMakholm As the LHS is $\approx 0.31$, I supposeit should be $0.2$ – Hagen von Eitzen Nov 20 '16 at 10:43
• Might be easier to rewrite the LHS to $\frac 22 H_{50} - H_{100} +1$. – Henning Makholm Nov 20 '16 at 10:44
• I can't understand why the OP doesn't answer direct questions asked to him. -1 – DonAntonio Nov 20 '16 at 10:49
• To DonAntonio actually it does not strike in my mind. – juantheron Nov 20 '16 at 10:50

$$\left(\frac12-\frac13\right)+\left(\frac14-\frac15\right)+\cdots+\left(\frac1{98}-\frac1{99}\right)+\frac1{100}>\left(\frac12-\frac13\right)+\left(\frac14-\frac15\right)=0.21\overline 6$$