Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying for my calculus exam and I'm stuck with the next exercise, I have to find out the monotonicity of the sequence $\{a_n\}$ with the next general term: $$a_n=\sum \limits_{p=1}^n\frac{1}{n+p}$$

For it to be monotone, the next condition must be proved: $a_n\le a_{n+1}$(incremetal) or $a_n\ge a_{n+1}$(decremental), so: $$\sum \limits_{p=1}^n\frac{1}{n+p}\le\sum \limits_{p=1}^{n+1}\frac{1}{(n+1)+p}$$


$$\sum \limits_{p=1}^n\frac{1}{n+p}\ge\sum \limits_{p=1}^{n+1}\frac{1}{(n+1)+p}$$

I'm stuck here, I don't know how I can operate this in order to verify if the sequence is monotone.

share|cite|improve this question
I hope "monotony" is not to be taken literally. Perhaps you meant "monotonicity?" – Chris Leary Jan 10 '13 at 21:21
@ChrisLeary Funny. The natural translation of the spanish word for monotonicty to english would be monotony, hence his mistake. – Git Gud Jan 10 '13 at 21:22
@GitGud Yep thats what happened, im not native english speaker hehe – Alejandro Jan 10 '13 at 21:24
Since the numbers $a_n$ are all positive, you can erase the absolute value signs, thereby making your life much more pleasant. (Actually, I'm not sure why those absolute value signs were there in the first place.) – Andreas Blass Jan 10 '13 at 21:33
You can see the monotonicty by writting out $a_n$ for a couple of terms. You'll see a pattern in the expanded summation. – AvatarOfChronos Jan 10 '13 at 21:38

Let $(p+1) = q$, $$a_{n+1} = \sum_{p=1}^{n+1}\frac{1}{n+1+p} = \sum_{q=2}^{n+2} \frac{1}{n+q} \\ = \sum_{q=1}^{n} \frac{1}{n+q} + \frac{1}{n + n+1}+\frac{1}{2n+2} - \frac{1}{n+1} = a_n + A$$

Now it remains to show that $A > 0$. From here, we get $A>0$ for $n \in \mathbb N$

Further more, we have $$\frac{1}{n + n+1}+\frac{1}{2n+2} - \frac{1}{n+1} > \frac{1}{2n+2}+\frac{1}{2n+2} - \frac{1}{n+1} = 0$$

And since we have $a_{n+1} > a_n$, the sequence $\{ a_n\}$is monotonically increasing.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.