Finding limit and sum of power series

Question

Have such limit:

$$ \lim \limits_{n \to \infty} \dfrac{1^{15}+3^{15}+ ...+ (2n-1)^{15}}{n^{16}} $$

But the sum

$$ \sum_{n=1}^\infty (2n-1)^{15} $$

diverges.

I think, that the answer is 0, because it's the sum of limits where the largest is 1/n but I can't prove it (if it is actually true). Can you help me to solve this?

Answer 1

Well using the Stolz-Cesaro theorem $$\lim \limits_{n \to \infty} \dfrac{1^{15}+3^{15}+ ...+ (2n-1)^{15}}{n^{16}}=\lim_{n \to \infty} \dfrac{(2n+1)^{15}}{(n+1)^{16}-n^{{16}}}=\frac{2^{15}n^{15}+\cdots1}{{16\choose 1}n^{15}+\cdots+1}=\frac{2^{15}}{2^4}=2^{11} $$ Clearly only the $n^{15}$ term is important that's why i just added dots from the binomial expansion,and ofc i just divided by $n^{15}$

Answer 2

Since $(2n-1)^{15}$ goes to infinity when $n\to \infty$, it's not the good road. However, $$a_n:=\frac 1{n^{16}}\sum_{j=1}^n(2j-1)^{15}=\frac 1n\sum_{j=1}^n\left(2\frac jn-\frac 1{n}\right)^{15}.$$ It's almost a Riemann sum, up to the term $1/n$, but we can see that if $b_n:= \frac 1n\sum_{j=1}^n\left(2\frac jn\right)^{15}$, then $\lim_{n\to +\infty}a_n-b_n=0$.

Answer 3

Here's a rather different approach. I will try to find the leading term of $a_n$ which I can then plug in the limit and hence find a solution.
Consider $a_n = \sum_{i=1}^ni^m$. Then I write the following difference equation : $$a_{n+1} - a_n = n^m$$ To solve this kind of equation, you normally let: $$a_n = c_{m+1}n^{m+1} + c_mn^m + ... + c_1n+c_0$$ Substituting in the difference equation, I only consider the two leading terms to find $c_{m+1}$: $$c_{m+1}(n+1)^{m+1} + c_m(n+1)^m - c_{m+1}n^{m+1} - c_mn^m + O(n^{m-1}) = n^m$$ Expand, the $n^{m+1}$ terms will cancel out and the $n^m$ terms will have coefficients satisfying the following equation: \begin{eqnarray*} c_{m+1} + mc_{m+1} &=& 1\\ \Longrightarrow c_{m+1} &=& \frac{1}{m+1}\end{eqnarray*}

This is the general case so in your case, you have $m = 15$ so let $$a_n = 1^{15} + 2^{15} + ... + n^{15} = \frac{1}{16}n^{16} + O(n^{15})$$
But the numerator, call it $t_n$, is only considering the odd numbers so you obtain :$$t_n = \underbrace{a_{2n-1}}_\text{sum from 1 to 2n-1} - \underbrace{2^{15}a_{n-1}}_\text{sum of all even from 2 to 2n-2} = \frac{1}{16}(2n-1)^{16} - \frac{1}{16}2^{15}n^{16} + O(n^{15})$$ Again, only considering the terms $n^{16}$, you can find : \begin{eqnarray*} t_n &=& \bigg(\frac{2^{16}}{16} - \frac{2^{15}}{16}\bigg)n^{16} + O(n^{15})\\ &=&2^{11}n^{16} + O(n^{15}) \end{eqnarray*}

Therefore the limit simplifies to $$\lim_{n\to \infty} \frac{t_n}{n^{16}} = 2^{11} = 2048$$