Prove $\sum^{\infty}_{n=1} \frac{a_{n}-a_{n-1}}{a_{n}}=\infty$

Question

Prove $$\sum^{\infty}_{n=1} \frac{a_{n}-a_{n-1}}{a_{n}}=\infty$$ Where $a_{n}$ is an increasing sequence of positive terms that goes to infinity.

I tried to approach it with $\log(a_{n})$ like a classical version of this problem but i could not show that the difference with respect to the logarithm is finite. Any help or hint will be appreciated

Consider that $a_n=n$ yields the harmonic series. You might try to reproduce an argument for that series with this one. — Thomas Andrews, Apr 09 '14 at 05:19
No, that wasn't what I was saying. There is a standard proof that $\sum_n \frac{1}{n}$ diverges. This theorem uses the same essential proof. — Thomas Andrews, Apr 09 '14 at 05:27
@user114032 strictly speaking but you can apply both arguments with a little to none adjustments. — clark, Apr 09 '14 at 14:00

score 3 · Answer 1 · answered Apr 09 '14 at 05:26

3

Show that:

$$\sum_{n=N+1}^M \frac{a_n-a_{n-1}}{a_n} \geq \frac{a_M-a_N}{a_M}$$

Then pick an increasing sequence $N_i$ so that $a_{N_{i+1}}\geq 2A_{N_i}$.

Then $$\sum_{n=N_i+1}^{N_{i+1}}\frac{a_n-a_{n-1}}{a_n}\geq 1$$

Use this to deduce the series diverges.

answered Apr 09 '14 at 05:26

Thomas Andrews

177,126

Nice, thanks a lot. Do you know if this can be compared to the log of an by any means. – user114032 Apr 09 '14 at 05:29
3

It is definitely a lower Reimann sum for $\int_{a_0}^{a_N}\frac{dt}{t}$, so the sums are bounded above by $\log(a_N/a_0)$. But I think it can diverge much more slowly... – Thomas Andrews Apr 09 '14 at 05:35
@ThomasAndrews why I can pick an increasing sequence $N_i$ so that $a_{N_{i+1}} \geq 2A_{N_i}$? – user10024395 Apr 22 '15 at 10:02

Prove $\sum^{\infty}_{n=1} \frac{a_{n}-a_{n-1}}{a_{n}}=\infty$

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