# The series $\sum_{n=1}^\infty \frac{n}{n+1}$

$$\sum_{n=1}^\infty \frac{n}{n+1}$$

I have $\displaystyle\lim_{n \to{+}\infty}{\frac{n}{n+1}}=\displaystyle\lim_{n \to{+}\infty}{\frac{\frac{n}{n}}{\frac{n}{n}+\frac{1}{n}}}=\displaystyle\lim_{n \to{+}\infty}{\frac{1}{1+\frac{1}{n}}}=1$

Since this is not 0, the series diverges by the divergence test.

Is it right to my result?

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That would depend on what the question is. If the question is "Does $\sum\frac{n}{n+1}$ converge?" then your answer is right. If the question is "Is $\sum\frac{n}{n+1}$ a series whose terms are all rational numbers?" then your answer would be wrong. Since you didn't tell us what the question was... –  Arturo Magidin May 22 '12 at 5:28

Here is another way of looking at it:

$$\sum_{n=1} ^{\infty} \frac{1}{n+1} \leq \sum_{n=1} ^{\infty} \frac{n}{n+1}$$

since $\frac{1}{n+1} \leq \frac{n}{n+1}$ for all $n>0$. Clearly $\sum_{n=1} ^{\infty} \frac{1}{n+1}$ diverges since it is the harmonic series minus $1$, and therefore $\sum_{n=1} ^{\infty} \frac{n}{n+1}$ diverges.

As t.b. points out in the comments there is a more basic way to bound this series. Notice that

$$\frac{1}{2}\leq \frac{n}{n+1}$$

for all positive $n$. Therefore

$$\sum_{n=1} ^{\infty} \frac{1}{2} \leq \sum_{n=1} ^{\infty} \frac{n}{n+1}.$$

Since $\sum_{n=1} ^{\infty} \frac{1}{2}$ is divergent, $\sum_{n=1} ^{\infty} \frac{n}{n+1}$ must diverge.

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Wouldn't it be much easier to say that $\frac{1}{2} \leq \frac{n}{1+n}$? –  t.b. May 22 '12 at 5:50
@t.b. I suppose it would. This was just my first thought. –  Holdsworth88 May 22 '12 at 5:53
Feel free to add that to your answer, if you do, I'd upvote it, because you're making a good point. –  t.b. May 22 '12 at 6:04
Thanks! ${}{}{}{}$ –  t.b. May 22 '12 at 6:20

Yes. You are right. However, there is no one universal divergence test. You need to say that since $\displaystyle \lim_{n \rightarrow \infty} a_n \neq 0$, we have that $\displaystyle \sum_{n=1}^{\infty} a_n$ doesn't converge.

EDIT As Arturo points out, looks like what I wrote above is "The divergence test". So you are fine provided the question is if the series diverges.

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And more to the point, the reasoning is sound. –  Harald Hanche-Olsen May 22 '12 at 5:27
@Marvis: In many calculus textbooks in the U.S., it is common to refer to that as "The Divergence Test." –  Arturo Magidin May 22 '12 at 5:29