Evaluate $\lim_{n \to \infty} \sum_{k=n}^{2n} \frac{1}{n+\sqrt{k}}$

Question

Evaluate $$\lim_{n \to \infty} \sum_{k=n}^{2n} \frac{1}{n+\sqrt{k}}$$

I tried to write the sum above as $$\lim_{n \to \infty} \frac{1}{n+\sqrt{n}} + \frac{1}{n+\sqrt{n+1}}+\dotsb+\frac{1}{n+\sqrt{2n}}$$ Now if I solve the limit the result is $0$ but obviously it's wrong. I think because it's an infinity sum of factors that go to zero, but it could be that the sum goes to infinity faster than the factors so that the result isn't zero. However I don't know how to solve it. Some advice?

Answer 1

$\frac{1}{n +\sqrt{k}} \leq \frac{1}{n}$, so the sum is upper-bounded by $\frac{1}{n} + \frac{1}{n} + \ldots \frac{1}{n} = 1 $

Answer 2

For $1\le n\le k\le2n$ we have

$$n+1\le n+\sqrt{k}\le n+\sqrt{2n}\le n+1+\sqrt{2n+2}$$

hence

$${1\over1+\sqrt{2\over n+1}}={n+1\over n+1+\sqrt{2n+2}}\le\sum_{k=n}^{2n}{1\over n+\sqrt k}\le{n+1\over n+1}=1$$

The Squeeze Theorem does the rest.

Answer 3

Wrote a small Python code to calculate this sum for specific $n$ values, to see where the limit of $n \to \infty$ goes:

import math

def limit(n):
    s = 0
    for i in range(n,2*n+1):
        s += 1/(n+math.sqrt(i))
    return s

for a in range(1,10001):
    print "n = " + str(a) + " | limit = " + str(limit(a))

print "n = 1000000 | limit = " + str(limit(1000000))
print "n = 10000000 | limit = " + str(limit(10000000))

Its output was

n = 1 | limit = 0.914213562373
n = 2 | limit = 0.810842411245
n = 3 | limit = 0.785811290103
n = 4 | limit = 0.778993834934
n = 5 | limit = 0.778479980466
n = 6 | limit = 0.780536268748
n = 7 | limit = 0.783702898
n = 8 | limit = 0.787331118896
n = 9 | limit = 0.791109376332
n = 10 | limit = 0.794881504095

$\vdots$

n = 9990 | limit = 0.988051600225
n = 9991 | limit = 0.988052185976
n = 9992 | limit = 0.988052771641
n = 9993 | limit = 0.988053357219
n = 9994 | limit = 0.988053942711
n = 9995 | limit = 0.988054528116
n = 9996 | limit = 0.988055113434
n = 9997 | limit = 0.988055698666
n = 9998 | limit = 0.988056283811
n = 9999 | limit = 0.988056868869
n = 10000 | limit = 0.988057453841
n = 1000000 | limit = 0.998783545517
n = 10000000 | limit = 0.99961478362

From this, it seems to very slowly approach $1$.

Answer 4

This sum is $$1+\frac{1}{n}-\sum_{k=n}^{2n}(\frac{1}{n}-\frac{1}{n+\sqrt{k}})\approx 1+\frac{1}{n}-\frac{1}{n^2}\sum_{k=n}^{2n}\sqrt{k}\approx 1+\frac{1}{n}-\frac{1}{n^2}\int_n^{2n}x^{-1/2}dx\in 1+\mathcal (\frac{1}{n}).$$

Answer 5

Use "sandwich theorem": $\displaystyle{n + 1 \over n + \,\sqrt{\,2n\,}\,} < \sum < {n + 1 \over n + \,\sqrt{\,n\,}\,}$.