Proving the Convergence: $\sum_{n=1}^{\infty}(\sqrt{n + r} - \sqrt{n})$.

Question

I'm having trouble proving the convergence for this series using real analysis methods: $$\sum_{n=1}^{\infty}a_n$$where for $r > 0$ a positive real number,$$a_n = \sqrt{n + r} - \sqrt{n}$$

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My approach so far has been to use the ratio test for $\frac{a_{k+1}}{a_k}$, which gives me the fraction $\frac{\sqrt{k + 1 + r} - \sqrt{k + 1}}{\sqrt{k + r} - \sqrt{k}}$. I multiply the top and bottom by the denominators conjugate, which conveniently leaves just $r$ on the bottom, but it leaves a pretty gross numerator: $$\frac{\sqrt{k^2 + 2kr + k + r^2 + r} + \sqrt{k^2 + kr + k} - \sqrt{k^2 + k + k + r} - \sqrt{k^2 + k}}{r}$$

At this point, a high school calculus style analysis would be content with me just saying that the limit of this is zero based on the degrees of the terms. But I'm totally lost on how I should prove that the limit for this is $0$ based on the $\epsilon$ definition for limits.

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Is there some sort of algebraic manipulation that we can do here to make the $\epsilon$ inequality for the limit definition turn out nicely? Alternatively, is there some other method I should be trying to use?

Thank you for your time and help:)

Answer 1

It is a divergent series. One has $$ \sum\limits_{n = 1}^\infty {(\sqrt {n + r} - \sqrt n )} = \sum\limits_{n = 1}^\infty {\frac{r}{{\sqrt {n + r} + \sqrt n }}} > \frac{r}{2}\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + r} }}} > \frac{r}{{2\sqrt {1+r} }}\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt n }}} , $$ and the last series is obviously divergent.

Answer 2

There are different ways to prove that the series diverges.

For natural $r$, the series telescopes, leaving you with

$$\lim_{n\to\infty}\sqrt{n+r}+\sqrt{n+r-1}+\cdots\sqrt{n+1}-f(r).$$

For non-natural $r$, you can use squeezing between the series for two naturals.

Answer 3

Let $n >n_0 >r$ ($n_0$ exists, Archimedean principle)

$a_n=\dfrac{r}{\sqrt{n+r}+√n}>$

$\dfrac{r}{2\sqrt{n+r}}>\dfrac{r}{2\sqrt{n+n}}=$

$\dfrac{r}{2√2√n}= \dfrac{r}{2√2}\dfrac{1}{√n}$.

$\sum a_n$ diverges (comparison test), since $\sum \dfrac{1}{√n}$ diverges

Answer 4

Since $r>0$ thus $\sqrt{n+r}-\sqrt{n}>0$. Let's find the $n$ for which $\sqrt{n+r}-\sqrt{n}>\frac{\alpha}{n}$ where $\alpha>0$ is some positive constant. Therefore

$$n+r+n-2\sqrt{n(n+r)}>\left( \frac{\alpha}{n} \right)^2\Rightarrow 2n+r-\left( \frac{\alpha}{n} \right)^2>2\sqrt{n(n+r)} \\ \Rightarrow 4n^2+r^2+\left( \frac{\alpha}{n} \right)^4+4nr-2r\left( \frac{\alpha}{n} \right)^2-4n\left( \frac{\alpha}{n} \right)^2>4n^2+4nr\\ \Rightarrow \left( r-\left( \frac{\alpha}{n} \right)^2 \right)^2>\frac{4\alpha^2}{n}\Rightarrow r>\left( \frac{\alpha}{n} \right)^2+\frac{2\alpha}{\sqrt{n}}$$

For large $n$ this is inevitable and the series is comparable to harmonic series which diverges. For example if we assume $n$ is large enough such that $\left( \frac{\alpha}{n} \right)<\left( \frac{\alpha}{\sqrt{n}} \right)$ (i.e. $n>1$) then

$$r+1>\left( \frac{\alpha}{\sqrt{n}}+1 \right)^2\Rightarrow \sqrt{r+1}>\left( \frac{\alpha}{\sqrt{n}} +1\right)\Rightarrow \sqrt{n}>\frac{\alpha}{\sqrt{r+1}-1}$$

Thus if $n>\left(\frac{\alpha}{\sqrt{r+1}-1}\right)^2$ then $\sum_{n=1}^{\infty}a_n>\sum_{n=1}^{\infty}\frac{\alpha}{n},\alpha>0$