Question

plsssss solve this question
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Answer 1

Given Appropriate Question :-

Prove that,

$\sf \: \displaystyle\lim_{n \to \infty }\bigg[\dfrac{ {1}^{2} }{ {1}^{3} + {n}^{3}} + \dfrac{ {2}^{2} }{ {2}^{3} + {n}^{3}} + - - - + \dfrac{ {n}^{2} }{ {n}^{3} + {n}^{3} }\bigg] = \frac{1}{3}log2$

Consider,

$\sf \: \displaystyle\lim_{n \to \infty }\bigg[\dfrac{ {1}^{2} }{ {1}^{3} + {n}^{3}} + \dfrac{ {2}^{2} }{ {2}^{3} + {n}^{3}} + - - - + \dfrac{ {n}^{2} }{ {n}^{3} + {n}^{3} }\bigg]$

can be rewritten as

$\rm \: = \:\displaystyle\lim_{n \to \infty }\displaystyle\sum_{r=1}^{r=n}\bigg[\dfrac{ {r}^{2} }{ {r}^{3} + {n}^{3} }\bigg]$

can be further rewritten as

$\rm \: = \:\displaystyle\lim_{n \to \infty }\displaystyle\sum_{r=1}^{r=n}\bigg[\dfrac{\dfrac{ {r}^{2} }{ {n}^{2} } \times \dfrac{1}{n} }{\dfrac{ {r}^{3} }{ {n}^{3} } + 1} \bigg]$

can be rewritten as

$\rm \: = \:\displaystyle\lim_{n \to \infty } \frac{1}{n} \displaystyle\sum_{r=1}^{r=n}\bigg[\dfrac{\dfrac{ {r}^{2} }{ {n}^{2} } }{\dfrac{ {r}^{3} }{ {n}^{3} } + 1} \bigg]$

We know,

Summation of series using Limit as sum is solved as

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty } \frac{1}{n} \displaystyle\sum_{r=h}^{r=g}f\bigg[\dfrac{r}{n} \bigg]$

is expressed as

$\rm \: = \:\displaystyle\int_{a}^{b}f(x) \: dx$

where,

$\boxed{ \tt{ \: \displaystyle\sum \: is \: replaced \: as \: \int}}$

$\boxed{ \tt{ \: \frac{r}{n} \: is \: replaced \: as \: x}}$

$\boxed{ \tt{ \: \frac{1}{n} \: is \: replaced \: as \: dx}}$

$\boxed{ \tt{ \: a = \displaystyle\lim_{n \to \infty } \frac{h}{n} }}$

$\boxed{ \tt{ \: b = \displaystyle\lim_{n \to \infty } \frac{g}{n} }}$

So, above expression can be rewritten as

$\rm \: = \:\displaystyle\int_{0}^{1} \frac{ {x}^{2} }{ {x}^{3} + 1} \: dx$

can be rewritten as

$\rm \: = \: \dfrac{1}{3} \displaystyle\int_{0}^{1} \frac{ 3{x}^{2} }{ {x}^{3} + 1} \: dx$

We know,

$\boxed{ \tt{ \: \displaystyle\int \: \frac{f'(x)}{f(x)} \: dx \: = \: logf(x) \: + \: c}}$

So, using this, we get

$\rm \: = \:\dfrac{1}{3} \bigg[log | {x}^{3} + 1| \bigg]_{0}^{1}$

$\rm \: = \:\dfrac{1}{3} \bigg[log | {1}^{3} + 1| - log |0 + 1| \bigg]$

$\rm \: = \:\dfrac{1}{3} \bigg[log |1 + 1| - log |1| \bigg]$

$\rm \: = \:\dfrac{1}{3} \bigg[log |2| \bigg]$

Hence, proved