23is the ans = x ?
\lim_{n\rightarrow \propto } \frac{\left[1^{3}x \right]+\frac{1}{2}\left[2^{3}x \right]+\frac{1}{3}\left[3^{3}x \right]+......+\frac{1}{n}\left[n^{3}x \right]}{1^{2}+2^{2}+3^{2}+...+n^{2}} , where [.] denotes greatest integer function.
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6 Answers
23r3x-1 < [r3x] ≤ r3x
divide by r
r2x - 1/r < [r3x]/r ≤ r2x
\Rightarrow \sum{r^{2}x-\frac{1}{r}} < \sum{\frac{[r^{3}x]}{r}} \leq \sum{r^{2}x}
\frac{\Rightarrow \sum{r^{2}x-\frac{1}{r}}}{\sum{r^{2}}} < \frac{\sum{\frac{[r^{3}x]}{r}}}{\sum{r^{2}}} \leq \frac{\sum{r^{2}x}}{\sum{r^{2}}}
x-\frac{\sum{\frac{1}{r}}}{\sum{r^{2}}} < \frac{\sum{\frac{[r^{3}x]}{r}}}{\sum{r^{2}}} \leq x
now
\frac{\sum{\frac{1}{r}}}{\sum{r^{2}}}=\frac{1+\frac{1}{2} +\frac{1}{3}+.....\frac{1}{n} }{\frac{n(n+1)}{2}}=\frac{1}{\frac{n(n+1)}{2}}+\frac{1}{2\frac{n(n+1)}{2}}+..\frac{1}{n\frac{n(n+1)}{2}}
\Rightarrow \lim_{n\rightarrow \infty}\frac{\sum{\frac{1}{r}}}{\sum{r^{2}}}\rightarrow 0
hence required limit = x
6its very simple
in infinity type probs
first try to present the quesn into N/D form
then divide N and D by highest power of x
u'll get the answer...