Proof that $\sum_{i=1}^n \frac{1}{n+i}$ is bounded

Question

$$\sum_{i=1}^n \frac{1}{n+i}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+(n-1)}+\frac{1}{2n}$$ I'm having problems proving that this sum is bounded. I know it's bounded from below because it's a sum of positive numbers but I'm having trouble proving it has an upper bound. I tried finding an upper bound for every part of the sum but I always end up with something depending on $n$ which diverges.

Answer 1

HINT

We have that $\forall n$

$$n\cdot \underbrace{\frac{1}{2n}}_{smaller}\le \overbrace{\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+(n-1)}+\frac{1}{2n}}^{n\,terms}\le n\cdot\underbrace{\frac{1}{n+1}}_{bigger}$$

Answer 2

Or let $$s=\sum_{i=1}^n \cfrac 1{n+i}$$ we have \begin{align} s^2 &<\left(\sum_{i=1}^n 1^2\middle)\middle(\sum_{i=1}^n \cfrac 1{(n+i)^2}\right)\hspace{2cm}\text{(Cauchy-Schwartz Inequality)}\\ &=n\sum_{i=1}^n \cfrac 1{(n+i)^2}\\ &<n\sum_{i=1}^n \cfrac 1{n(n+i)}\\ &=n\left(\cfrac 1n-\cfrac 1{n+1}+\cfrac 1{n+1}-\cfrac 1{n+2}+\cdots+\cfrac 1{2n-1}-\cfrac 1{2n}\right) \hspace{2mm}\text{(telescope series)}\\ &=n(\cfrac 1n-\cfrac 1{2n})=\cfrac 12 \end{align}

Hence $s< \cfrac {\sqrt{2}} 2$ and has to be bounded.

Answer 3

To prove that this series is bounded by an upper limit, would it suffice to prove that it is less than some other finite sum? For every term, $1/(n+i) <1/(n)$. On summing up to $n$ terms , the given sum is $<n*(1/(n))$, i.e. it is $<1$. So it is bounded. Please correct me if I am wrong.