Asymptotic for the following series.

Question

$$S_N=\frac{\sum_{i=1}^{2N}\sqrt{i}+\sqrt{i-1}}{\sum_{i=1}^{N}\sqrt{i}+\sqrt{i-1}}$$ I want to find an equivalence/asymptotic for $S_N$ as $N$ become very large. I tried the following:

Edit

We know that $$\sum_{i=1}^{N}\sqrt{i}\approx\frac{2N^{3/2}}{3}$$ and so $$S_N\approx\frac{(2N)^{3/2}+(2N-1)^{3/2}}{N^{3/2}+(N-1)^{3/2}}\approx 2\sqrt{2}.$$ Is this estimation correct and also how can it be improved?

Answer 1

Your formula is correct apart from the final simplification, which should be: $$ \frac{(2N)^{3/2}+(2N-1)^{3/2}}{N^{3/2}+(N-1)^{3/2}}\approx2^{3/2}. $$ To see this, observe that the two instances of $-1$ become irrelevant as $N\to\infty$, as long as you are looking for leading order asymptotics.

Answer 2

Here is a reasonably elementary proof for an arbitrary exponent, not just $\frac12$. It shows that if the exponent is $a$ with $a > 0$, then the limit is $2^{a+1}$. It also gives explicit bounds.

It is based on this:

If $a > 0$ then $\dfrac{n^{a+1}}{a+1}+n^a \gt \sum_{1}^{n} k^a \gt \dfrac{n^{a+1}}{a+1} $.

(Similar bounds can be gotten for $a < 0$.)

Proof:

$(k-1)^a < \int_{k-1}^{k} x^a dx \lt k^a $ so $\sum_{k=1}^n (k-1)^a \lt \sum_{k=1}^n \int_{k-1}^{k} x^a dx \lt \sum_{k=1}^nk^a $ or $\sum_{k=0}^{n-1} k^a \lt \int_{0}^{n} x^adx \lt \sum_{1}^{n} k^a $ so that, since $\int_{0}^{n} x^adx =\dfrac{n^{a+1}}{a+1} $, $-n^a \lt \dfrac{n^{a+1}}{a+1}-\sum_{1}^{n} k^a \lt 0 $. or $\dfrac{n^{a+1}}{a+1}+n^a \gt \sum_{1}^{n} k^a \gt \dfrac{n^{a+1}}{a+1} $.

Let $p(n) =\sum_{k=1}^n (k^a+(k-1)^a) $ and $S(n) =\dfrac{p(2n)}{p(n)} $.

$\begin{array}\\ p(n) &=\sum_{k=1}^n (k^a+(k-1)^a)\\ &=\sum_{k=1}^n k^a+\sum_{k=1}^n(k-1)^a\\ &=\sum_{k=1}^n k^a+\sum_{k=0}^{n-1}k^a\\ &=\sum_{k=1}^n k^a+\sum_{k=1}^{n-1}k^a\\ &=2\sum_{k=1}^n k^a-n^a\\ \end{array} $

so that $2\dfrac{n^{a+1}}{a+1}-n^a \lt p(n) \lt 2\dfrac{n^{a+1}}{a+1}+2n^a $.

Therefore

$\begin{array}\\ S(n) &\lt \dfrac{2\dfrac{(2n)^{a+1}}{a+1}+2(2n)^a}{2\dfrac{n^{a+1}}{a+1}-n^a}\\ &= \dfrac{ 2^{a+2}+(a+1)2^{a+1}/n}{2-(a+1)/n}\\ &= \dfrac{ 2^{a+1}+(a+1)2^{a}/n}{1-(a+1)/(2n)}\\ \end{array} $

and

$\begin{array}\\ S(n) &\gt \dfrac{2\dfrac{(2n)^{a+1}}{a+1}-(2n)^a}{2\dfrac{n^{a+1}}{a+1}+2n^a}\\ &= \dfrac{ 2^{a+2}-2(a+1)2^{a+1}/n}{2+2(a+1)/n}\\ &= \dfrac{ 2^{a+1}-2(a+1)2^{a}/n}{1+2(a+1)/n}\\ \end{array} $

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