Another limit

Kaymant sir ... what's the ans ?????

Nice proof by Bhatt sir. I had a different approach in mind. We shall prove the following general result:
Suppose f be a differentiable function with a continuous derivative on [0,1], then
\lim_{n\to\infty}\ n\left(\dfrac{1}{n}\sum_{k=1}^n f\left(\dfrac{k}{n}\right)-\int_0^1 f(x)\ \mathrm dx\right)=\dfrac{f(1)-f(0)}{2}

For the current problem, we have f(x)=x^\alpha. Accordingly, the required limit is
\dfrac{f(1)-f(0)}{2}=\dfrac{1}{2}

To prove the above assertion, we start out by writing
\int_0^1 f(x) \ \mathrm dx=\sum_{k=1}^n\int_{\frac{k-1}{n}}^{\frac{k}{n}} f(x)\ \mathrm dx
Also
\dfrac{1}{n}=\int_{\frac{k-1}{n}}^{\frac{k}{n}}\mathrm dx
So we get
\dfrac{1}{n}\sum_{k=1}^nf\left(\dfrac{k}{n}\right)-\int_0^1f(x)\ \mathrm dx =\sum_{k=1}^n\int_{\frac{k-1}{n}}^{\frac{k}{n}}\left(f\left(\dfrac{k}{n}\right)-f(x)\right)\mathrm dx
Next, use the LMVT to write
f\left(\dfrac{k}{n}\right)-f(x)=f'(c_k)\left(\dfrac{k}{n}-x\right)
where c_k\in\left(x,\frac{k}{n}\right).
Next, since f'(x) is continuous, it attains a minimum and maximum in every closed interval. Let
m_k = the minimum value of f'(x) in the interval [k-1n, kn], and
M_k = the maximum value of f'(x) in the interval [k-1n, kn]
Then, m_k ≤ f'(c_k) ≤ M_k for k = 1, 2, . . ., n
Hence, we get
m_k\int_{\frac{k-1}{n}}^{\frac{k}{n}}\left(\frac{k}{n}-x\right)\mathrm dx\le \int_{\frac{k-1}{n}}^{\frac{k}{n}}\left(f\left(\frac{k}{n}\right)-f(x)\right)\mathrm dx\le M_k\int_{\frac{k-1}{n}}^{\frac{k}{n}}\left(\frac{k}{n}-x\right)\mathrm dx
That is \dfrac{m_k}{2n^2}\le \int_{\frac{k-1}{n}}^{\frac{k}{n}}\left(f\left(\frac{k}{n}\right)-f(x)\right)\mathrm dx\le \dfrac{M_k}{2n^2}
Hence,
\dfrac{1}{2n^2}\sum_{k=1}^nm_k\le \sum_{k=1}^n\int_{\frac{k-1}{n}}^{\frac{k}{n}}\left(f\left(\frac{k}{n}\right)-f(x)\right)\mathrm dx\le \dfrac{1}{2n^2}\sum_{k=1}^nM_k
Hence,
\dfrac{1}{2n^2}\sum_{k=1}^nm_k\le \dfrac{1}{n}\sum_{k=1}^nf\left(\dfrac{k}{n}\right)-\int_0^1f(x)\ \mathrm dx \le \dfrac{1}{2n^2}\sum_{k=1}^nM_k
And therefore
\dfrac{1}{2n}\sum_{k=1}^nm_k\le n\left(\dfrac{1}{n}\sum_{k=1}^nf\left(\dfrac{k}{n}\right)-\int_0^1f(x)\ \mathrm dx\right) \le \dfrac{1}{2n}\sum_{k=1}^nM_k
Now as n→ ∞,
\dfrac{1}{2n}\sum_{k=1}^nm_k\to \dfrac{1}{2}\int_0^1f'(x)\ \mathrm dx
and
\dfrac{1}{2n}\sum_{k=1}^nM_k\to \dfrac{1}{2}\int_0^1f'(x)\ \mathrm dx
Hence, by the sandwich principle,
\boxed{\lim_{n\to\infty}n\left(\dfrac{1}{n}\sum_{k=1}^nf\left(\dfrac{k}{n}\right)-\int_0^1f(x)\ \mathrm dx\right)=\dfrac{1}{2}\int_0^1f'(x)\ \mathrm dx=\dfrac{f(1)-f(0)}{2}}
as desired.

thank u sir [1]

You can use the fact that \sum_{k=1}^n k^m = P(n) where P(n) is a polynomial of degree m+1 for natural m.

Further the coefficient of n^m+1 is \frac{1}{m+1} and the coefficient of n^m is always \frac{1}{2}

These facts are easily proved using induction and the standard ways of deriving the sum of powers of the 1st n natural numbers. (You can also consult http://en.wikipedia.org/wiki/Faulhaber's_formula or read up on Bernoulli numbers.

So P(n) = \frac{n^{m+1}}{m+1} + \frac{n^m}{2} + Q(n) where Q(n) is of degree m-1.

Now, we can evaluate the limit for natural numbers m.

We can write the limit as

\lim_{n \rightarrow \infty} n \left( \frac{\frac{n^{m+1}}{m+1} + \frac{n^m}{2} + Q(n)}{n^{m+1}} - \frac{1}{m+1} \right) = \frac{1}{2}

Now for any general real numbers α>1. Let [α] = m.

We have

\sum_{k=1}^n \left(\frac{k}{n} \right)^{m+1} < \sum_{k=1}^n \left(\frac{k}{n} \right)^{\alpha} \le \sum_{k=1}^n \left(\frac{k}{n} \right)^{m}

Hence the required limit can be sandwiched between the limits evaluated for α=m+1 and α=m. Since from the above discussion both the left limit and right limit evaluate to 1/2, the required limit by Sandwich Theorem or Squeeze Theorem is \boxed { \frac{1}{2}}

It's alright ... every body makes errors ...

sorry its not correct i made a small error

ans is \frac{\alpha }{\alpha +1}?

an attempt :
\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\frac{1}{n}{\left(\frac{r}{n} \right)^{\alpha}}=\lim_{n\rightarrow \infty}\int_{\frac{1}{n}}^{1}{x^{\alpha}\mathrm{dx}} \\ \frac{x^{\alpha+1}}{1+\alpha}|_{\frac{1}{n}}^{1}=\frac{1}{1+\alpha}\left( 1-\frac{1}{n^{\alpha +1}}\right) \\ \frac{1}{1+\alpha}\lim_{n\rightarrow \infty}n(\frac{1}{n^{1+\alpha}})=0

Hey Soumika ... check Sonne's 2nd step ....

The solution stands ...

1α+1(1 - 1n^α+1)

Hence ... The expr. ultimately stands ....

EXP. = 1/(α+1) - 1/{(α+1)(n^α+1)} * n ... as given earlier

Now [ EXP. - n/(α+1) ] = -n/{(α+1)(n^α+1)}

Since n is natural no. ... therefore the new expr. is < 0.

This is my logic .. might be wrong ... not sure ...

Here lies the real problem ...

if now n -> ∞ .. the final expr. that I derived is -∞ which u got and so do I ...

is the answer -∞

Sir is it ZERO??