convergence divergence - Computing terms of a sequence and proving it's convergent

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

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.