5
$\begingroup$

I have to proof that $\sum\limits_{n=1}^{\infty}\frac{5}{10+n}$ either converges or diverges.

The first thing that I noticed is that it looks like the divergent harmonic series. First I did the nth term divergence test, which turned out to be inconclusive.

Then, because of the aforementioned harmonic series, I thought I'd do a comparison test. If $\sum\limits_{n=1}^{\infty}\frac{1}{n} \leq \sum\limits_{n=1}^{\infty}\frac{5}{10+n}$, and we know the harmonic series diverges, well then $\sum\limits_{n=1}^{\infty}\frac{5}{10+n}$ diverges. Now, here's the thing:

I could write $\sum\limits_{n=1}^{\infty}\frac{5}{10+n}$ as $5\cdot\sum\limits_{n=1}^{\infty}\frac{1}{10+n}$, but $\frac{1}{n} \nleq \frac{1}{10+n}$. However, $\frac{1}{n} \leq \frac{5}{10+n}$, but only from $n \geq 3$.

Here's what I don't know what to do:In all the examples I've seen, $n$ is always $\geq 1$, and there is no mention if it matters if you start your $n$ from somewhere else.

Second thing is, it is easy to see that $\frac{1}{n} \nleq \frac{1}{10+n}$, since a bigger denominator means a smaller number. However, with $\frac{1}{n} \leq \frac{5}{10+n}$ it isn't that easy to see. How could I proof it is so, from $n \geq 3$?

Third option ;-): Just do a limit comparison test!

$\endgroup$
  • $\begingroup$ You can do the limit comparison test or the integral test. $\endgroup$ – Joe Johnson 126 Apr 2 '15 at 10:09
  • $\begingroup$ Yeah, that's what I went for in the end, the limit comparison test. Just wondered if I should bother figuring it out for the comparison test. $\endgroup$ – Garth Marenghi Apr 2 '15 at 10:10
  • $\begingroup$ Leaving out the factor $5$ (which is irrelevant for convergence), you can add to the series the terms $1+1/2+\dots+1/10$ and … $\endgroup$ – egreg Apr 2 '15 at 10:12
  • 1
    $\begingroup$ Let $n$ be positive. Then $1/n \le 5/(10+n) \Leftrightarrow 10+n \le 5n \Leftrightarrow 10 \le 4n \Leftrightarrow 5/2 \le n$. $\endgroup$ – user37238 Apr 2 '15 at 10:20
  • $\begingroup$ @GarthMarenghi By the way, the question shows clearly that you work on the question of the convergence before asking here a question. Thank you. But you don't clearly (in my opinion) state what is your problem. $\endgroup$ – user37238 Apr 2 '15 at 11:22
2
$\begingroup$

You can use direct comparison test (my personal favorite when I can make it work). One thing to keep in mind is that the inequality does not have to be true for all $n$, just the entire "tail" of the series. This is because the first finitely many terms always sum to something finite. It only matters what happens for all $n\geq N$ (you can specify $N$ to be any finite number you like).

You could do a direct comparison like this (so that the inequality is easy to see): Keep in mind that $\displaystyle C\sum_{n=1}^{\infty}\frac{1}{n}$ diverges for any $C>0$, in this case, I will pick $C=\frac{1}{11}$.

$$\frac{1}{11}\sum_{n=1}^{\infty}\frac{1}{n}=\sum_{n=1}^{\infty}\frac{1}{11n} \leq \sum_{n=1}^{\infty} \frac{1}{n+10}\leq \sum_{n=1}^{\infty} \frac{5}{n+10}.$$

Note that $11n \geq n+10$ since $n\geq 1$ and smaller denominator (same numerator) means bigger fraction.

An alternative might be this: Notice that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n+10}$ is the same as $\displaystyle \sum_{n=11}^{\infty} \frac{1}{n}$. Since $\displaystyle \sum_{n=1}^{10}\frac{1}{n}$ is finite and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ is infinite, you may conclude that $\displaystyle \sum_{n=11}^{\infty}\frac{1}{n}$ is also infinite (otherwise, the sum of two finite numbers cannot be infinite).

$\endgroup$
1
$\begingroup$

Another type of comparison test shows that

$\frac{ \frac{1}{n} }{\frac{5}{10+n}} \to \frac{1}{5}$(const.) as $n\to \infty$ and $\sum^\infty \frac{1}{n}\to \infty$ as $n\to \infty$ implies $\sum^\infty \frac{5}{10 + n}$ diverges.

$\endgroup$
  • $\begingroup$ This looks like the limit comparison test (although there, $\frac{1}{n}$ would be the denominator). Yeah, this works too. $\endgroup$ – Garth Marenghi Apr 2 '15 at 13:03
0
$\begingroup$

In fact, the property $\sum_{n=0}^\infty ra_n=r\sum_{n=0}^\infty a_n$ is applied only when the given series $\sum_{n=0}^\infty a_n$ is converge, so it does not make sense when you write $\sum_{n=1}^\infty\frac{5}{10+n}$ as $5.\sum_{n=1}^\infty\frac{1}{10+n}$.

$\endgroup$
  • $\begingroup$ I was not aware of that. So, in essence, I always compare to $\frac{5}{10+n}$ instead of $\frac{1}{10+n}$. $\endgroup$ – Garth Marenghi Apr 2 '15 at 11:12
0
$\begingroup$

EVERY series either converges or diverges, so there's nothing to prove here... Equivalently, the proof is straightforward.

However if you have to determine whether the series converges or diverges and prove that, I would suggest to substitute $k=10+n$ and look at the resulting series $$\sum\limits_{k=11}^{\infty}\frac5k$$ which is $$\sum\limits_{k=11}^{\infty}\frac5k = 5\sum\limits_{k=11}^{\infty}\frac1k = 5\left( \sum\limits_{k=1}^{\infty}\frac1k - \sum\limits_{k=1}^{10}\frac1k\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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