6.0 Summation of Series Using Definite Integral

$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = {r_1}}^{{r_2}} {{1 \over n}f\left( {{r \over n}} \right) = \int\limits_{\mathop {\lim }\limits_{n \to \infty } {{{r_1}} \over n}}^{\mathop {\lim }\limits_{n \to \infty } {{{r_2}} \over n}} {f\left( x \right)dx} } $$

It is a very important formula and can be used to solve many Questions including limit with series.

Question 8. $\mathop {\lim }\limits_{n \to \infty } \left( {{1 \over n} + {1 \over {n + 1}} + ... + {1 \over {4n}}} \right)$

Solution: $$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{3n} {{1 \over {n + r}}} $$ $$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{3n} {{1 \over n}{1 \over {1 + {r \over n}}}} $$ $$ = \int\limits_{\mathop {\lim }\limits_{n \to \infty } {0 \over n}}^{\mathop {\lim }\limits_{n \to \infty } {{3n} \over n}} {{1 \over {1 + x}}dx} $$ $$ = \int\limits_0^3 {{1 \over {1 + x}}dx} $$ $$ = \left[ {\ln \left( {1 + x} \right)} \right]_0^3$$ $$ = \ln \left( {1 + 3} \right) - \ln \left( {1 + 0} \right)$$ $$ = \ln 4 - \ln 1$$ $$ = \ln 4$$

Question 9. $S = \mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {1 + \sqrt n }} + {1 \over {2 + \sqrt {2n} }} + ... + {1 \over {n + \sqrt {{n^2}} }}} \right]$

Solution: $$S = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over {r + \sqrt {rn} }}} $$ $$S = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{1 \over {\left( {{r \over n} + \sqrt {{r \over n}} } \right)}}} $$ $$S = \int\limits_0^1 {{1 \over {x + \sqrt x }}dx} $$ $$S = \int\limits_0^1 {{1 \over {\sqrt x \left( {\sqrt x + 1} \right)}}dx} $$

Let $\sqrt x + 1 = t$ then $${1 \over {2\sqrt x }}dx = dt$$

As x varies from $0$ to $1$, $t$ varies from $1$ to $2$ so limit changes.$$S = \int\limits_1^2 {{2 \over t}dt} $$ $$S = \left[ {2\ln t} \right]_1^2$$ $$S = 2\ln 2 - 2\ln 1$$ $$S = 2\ln 2$$