Computing terms of a sequence and proving it's convergent

Question

Let $c_n$ be the sequence defined by $$c_n =1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} -2\sqrt n$$

a) Compute $c_1$, $c_2$, and $c_3$.

b)Prove that $c_n$ is a convergent sequence.

My attempt was:

a) $c_1$ = $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{1}} -2\sqrt 1$ = $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + 1 - 2$ = $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots$

and same thing for $c_2$ and $c_3$ by replacing n by 2 and 3 respectively. Concerning part b, I don't know how it's done... any help?

Answer 1

We may write $$ c_n =\sum_{k=1}^n f(k) -\int_0^n f(x)\mathrm dx $$ where $f(x) = \frac{1}{\sqrt{x}},\ x> 0$. Note that since $f$ is decreasing, we have for every $k\ge 2$, $$ f(k-1)\ge \int_{k-1}^k f(x)\mathrm dx \ge f(k), $$ which implies $$ \sum_{k=1}^{n-1} f(k)\ge \int_1^n f(x)\mathrm dx \ge \sum_{k=2}^{n} f(k) $$ by summing over $2\le k\le n$. This in turn implies that $$ c_n =\sum_{k=1}^{n-1} f(k)-\int_1^n f(x)\mathrm dx +\frac{1}{\sqrt{n}}-2\ge -2 $$ and $$ c_n-c_{n+1} = \int_n^{n+1} f(x)\mathrm dx -f(n+1)\ge 0 $$ for every $n\ge 1$. As a result, bounded decreasing sequence $(c_n)$ has a finite limit $\lim\limits_{n\to\infty} c_n$ by monotone convergence theorem.

Answer 2

Hint:

$$c_n-c_{n+1}=2\sqrt{n+1}-2\sqrt{n}-\frac{1}{(n+1)^{1/2}}=2\sqrt{n}\left(\sqrt{1+\frac{1}{n}}-1\right)-\frac{1}{(n+1)^{1/2}}=2\sqrt{n}(n^{-1}/2+O(n^{-2}))-(n+1)^{-1/2}=-n^{-1/2}((1+1/n)^{-1/2}-1)+O(n^{-3/2})=O(n^{-3/2})$$

Answer 3

First observe that $c_n - c_{n+1} = \frac {1} {(\sqrt {n} + \sqrt {n+1})^2 \sqrt {n+1}} > 0$ for all $n \geq 1$. So the sequence $\{c_n \}$ is monotone decreasing.Also observe by recursion that $c_n = c_1 - \sum\limits_{k=1}^{n-1} \frac {1} {(\sqrt {n} + \sqrt {n+1})^2 \sqrt {n+1}}$ for all $n \geq 1.$ Since $\frac {1} {(\sqrt {n} + \sqrt {n+1})^2 \sqrt {n+1}} < \frac {1} {n^{\frac 3 2}}$ for all $n \geq 1$ the series $\sum\limits_{n=1}^{\infty} \frac {1} {(\sqrt {n} + \sqrt {n+1})^2 \sqrt {n+1}}$ is convergent. Let the sum be $S.$ Then we have $c_n \geq - (S+1)$ for all $n \geq 1.$ Hence the sequence $\{c_n \}$ is bounded below by $-(S+1).$ As the sequence $\{c_n \}$ is monotone decreasing and bounded below it is convergent.