1Kaymant sir ... what's the ans ?????
Find the following limit for \alpha >1:
\lim_{n\to \infty} n\left(\dfrac{1^\alpha + 2^\alpha + \ldots + n^\alpha}{n^{\alpha+1}}-\dfrac{1}{\alpha +1}\right)
Here n is a natural number.
-
UP 0 DOWN 0 2 20
20 Answers
66Nice 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
mk = the minimum value of f'(x) in the interval [k-1n, kn], and
Mk = the maximum value of f'(x) in the interval [k-1n, kn]
Then, mk ≤ f'(ck) ≤ Mk 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.
341You 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 nm+1 is \frac{1}{m+1} and the coefficient of nm 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}}
1thats wrong
the last few terms will be 1
like lim n trnds to infinity n-1/n =1 as well and not only last term
1an 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
1Hey Soumika ... check Sonne's 2nd step ....
The solution stands ...
1α+1(1 - 1nα+1)
Hence ... The expr. ultimately stands ....
EXP. = 1/(α+1) - 1/{(α+1)(nα+1)} * n ... as given earlier
Now [ EXP. - n/(α+1) ] = -n/{(α+1)(nα+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 ...
1I don't understand where it is wrong ... I'm getting the same answer by using sum of an integral...