Question

Show that lim n tends to infinity [1/n + 1/n+1 + 1/n+2 +...+ 1/3n] = log 3

Answer 1

Concept:

Limit is used to give values to functions which does not have defined values. It is used in mathematical physics.

Given:

The equation $\lim_{n \to \infty}[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2} +......+\frac{1}{3n}]=\log3$

Find:

Proof the equation.

Solution:

According to the problem,

$S= \lim_{n \to \infty}[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2} +......+\frac{1}{3n}]$

$S= \lim_{n \to \infty} [\frac{1}{n}\sum^{2n}_{r=0}\frac{1}{n+r} ]$

$S= \lim_{n \to \infty} [\frac{1}{n}\sum^{2n}_{r=0}\frac{n}{n+r} ]$

$= \lim_{n \to \infty} [\frac{1}{n}\sum^{2n}_{r=0}\frac{1}{1+(\frac{r}{n}) } ]$

Now lower the limit $\lim_{n \to \infty} \frac{r}{n}=0$

Since, $r=0$ for the first term.

Upper limit $\lim_{n \to \infty} \frac{r}{n}$

Since, $r=2n$ for the last term.

$S= \lim_{n \to \infty} [\frac{1}{n}\sum^{2n}_{r=0}\frac{1}{1+\frac{r}{n} } ]$

$S= \int\limits^2_0 {\frac{1}{x} } \, dx$

$=[\log(1+x)]^2_0$

$=\log3$

Thus $L.H.S=R.H.S$

Hence, when limit $n$ tends to infinity when the series becomes $\log3$ proofing the equation.