Could I use telescoping sum?

Calculus Level 4

$\large \lim_{x \to 0} \left (1^{\frac 1 {\sin^2 x}} + 2^{\frac 1 {\sin^2 x}} + 3^{\frac 1 {\sin^2 x}} + \ldots + n^{\frac 1 {\sin^2 x}} \right )^{\sin^2 x } = \ ?$

\infty

n

1

None of these choices

\frac{n+2}{2}

0

3 solutions

Tanishq Varshney
Apr 29, 2015

I think this is direct application of Cauch'theorem

Nayan Pathak - 6 years, 1 month ago

You have not provided an upper bound.

Harsh Shrivastava - 3 years, 6 months ago

Aditya Kumar
May 19, 2015

For large powers $n^a>>(n-1)^a$ . Therefore, the rest of the terms can be neglected. then the answer is left with $n$ . This is a worthy method for jee, taught to me by Z.Ahmed (the solver of Ahmed's integral)

Curtis Clement
Apr 28, 2015

Tanishq please can you hint as to how the limit equals n

I did it this way. Put $y=\frac {1}{sin^2(x)}$ . Now, $y \to \infty$

Let:

$a(n)=lim_{y \to \infty} (1^y+2^y+...+n^y)^{1/y}$

Observe:

$a(n)=lim_{y \to \infty} n\times(1+(\frac {a(n-1)} {n})^y )^{1/y}$

Since, $a(1)=1$ , $a(n)=n$ by induction.

Raghav Vaidyanathan - 6 years, 1 month ago

@Curtis Clement

Raghav Vaidyanathan - 6 years, 1 month ago

how about using sandwich theorem

Tanishq Varshney - 6 years, 1 month ago

@Tanishq Varshney – Sounds tasty :P

Curtis Clement - 6 years, 1 month ago

Could I use telescoping sum?

3 solutions

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