$a_{n+1}-a_{n}\ge\frac{1}{n}$. Prove: $\lim\limits_{n \to \infty} a_n= \infty$

Question

$\{a_n\}$ a sequence such that for all $n$:

$a_{n+1}-a_{n}\ge\frac{1}{n}$

Prove: $\lim\limits_{n \to \infty} a_n= \infty$

I have figured already that $a_{n+1}-a_{1}\ge\sum_{k=1}^n \frac{1}{k}$. Am I on the right track?

Answer 1

I suppose that what you proved is that$$a_{n+1}-a_1\geqslant\sum_{k=1}^n\frac1k.$$Now, use the fact that the harmonic series diverges.

Answer 2

As a note, the original post basically does it, except for just three observations:

• $\lim_{n \rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} a_{n}$ (if the limit exists)
• You have already shown $a_{n+1} \geq a_1 + \sum_{k=1}^n \frac{1}{k}$
• The right-hand series diverges to infinity for large $n$.

I suspect the only real missing step you were "stuck" on was the first and easiest obvervation!

If you need proof of the last observation, note that $$\sum_{k=1}^n \frac{1}{k} = \sum_{p=0}^{n}\sum_{m = 2^p}^{2^{p+1} - 1} \frac{1}{m} > \sum_{p=0}^{n}\sum_{m = 2^p}^{2^{p+1} - 1} \frac{1}{2^{p+1}} = \sum_{p=0}^{n} \frac{(2^{p+1} - 1) - 2^p}{2^{p+1}} = \sum_{p=0}^{n} \left(\frac{- 1}{2^{p+1}} + \frac{1}{2} \right) \\= \frac{n+1}{2} - \sum_{p=0}^{n} \frac{1}{2^{p+1}} = \frac{n+1}{2} - \frac{1}{2}\sum_{p=0}^{n} \left(\frac{1}{2}\right)^p > \frac{n+1}{2} - \frac{1}{2}( 2 ) = \frac{n-1}{2} $$

and the last term is positive linear in $n$, so it obviously diverges to infinity!