Limits and series

Calculus Level 3

Let the r r th term t r t_r of a series given by t r = r 1 + r 2 + r 4 t_r = \dfrac{r}{1+r^2+r^4} .The find the value of lim n r = 1 n t r \lim_{n\rightarrow \infty} \displaystyle\sum_{r=1}^n t_r


The answer is 0.5.

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1 solution

Rishabh Jain
Feb 25, 2016

S = r = 1 n r 1 + r 2 + r 4 = r = 1 n r ( r 2 + 1 r ) ( r 2 + 1 + r ) \large \mathfrak S=\displaystyle \sum_{r=1}^{n}\dfrac{r}{1+r^2+r^4}\\\large= \displaystyle \sum_{r=1}^{n}\dfrac{r}{(r^2+1-r)(r^2+1+r)} = 1 2 ( r = 1 n 1 r 2 + 1 r 1 r 2 + 1 + r ) ( A Telescopic series) \large =\dfrac{1}{2}(\displaystyle \sum_{r=1}^{n}\dfrac{1}{r^2+1-r} - \dfrac{1}{r^2+1+r})\\ (\color{#0C6AC7}{\small{\text{A Telescopic series)}}} = 1 2 ( 1 1 n 2 + n + 1 ) \large =\dfrac{1}{2}(1-\dfrac{1}{n^2+n+1}) = 1 2 ( n ( n + 1 ) ( n 2 + n + 1 ) ) \large=\dfrac{1}{2}(\dfrac{n(n+1)}{(n^2+n+1)}) As n , S approaches 0.5 \Large\text{As }n\rightarrow {\infty},~\mathfrak{S} \text{ approaches}\huge\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{\textbf{0.5}}}}}}

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