# Math Help - Converging sequenses

1. ## Converging sequenses

Determine which of the following sequences, , converge and find the limit of those that do

i)

ii)

2. Originally Posted by matty888
Determine which of the following sequences, , converge and find the limit of those that do

i)
If $\lim_{n\to\infty} a_n$ does not exist or if $\lim_{n\to\infty} a_n \not{=} 0$, then the series $\sum_{n=1}^\infty a_n$ is divergent.

(this should be rather obvious, if you are continuing to add a number infinity times, and it is not zero, it will diverge towards infinity)

3. Originally Posted by matty888
ii)
Using the Root test

$\lim_{n\to\infty} \sqrt[n]{|a_n|} = \lim_{n\to\infty} \sqrt[n]{|(n^2+n)^{1/n}|} = \lim_{n\to\infty} n^2+n$

Which diverges towards infinity, which is greater than one, and so the series is therefore divergent.

4. Hello,

Is this a new method ?

$\sqrt[n]{a}=a^{\frac{1}{n}}$

So here $\lim_{n\to\infty} \sqrt[n]{|(n^2+n)^{1/n}|}=\lim_{n\to\infty} (n^2+n)^{1/n^2}$

5. Originally Posted by Moo
Hello,

Is this a new method ?

$\sqrt[n]{a}=a^{\frac{1}{n}}$

So here $\lim_{n\to\infty} \sqrt[n]{|(n^2+n)^{1/n}|}=\lim_{n\to\infty} (n^2+n)^{1/n^2}$
Crap >.<

6. Originally Posted by matty888
Determine which of the following sequences, , converge and find the limit of those that do

i)

ii)
ok this is fairly simple

i) $a_n=\frac{n^2}{15n+5}$...so we need to take the limit to see if it diverges by the n-th term test...o wait we just want diverging sequences so we will take the limit and see if we get a non-∞ answer...so we have $\lim_{n \to {\infty}}\frac{n^2}{15n+5}$..multiplying top and bottom by $\frac{\frac{1}{n}}{\frac{1}{n}}$ or using L'hopitals rule the answer is obviously ∞ which means the series is divergent...

ii) we have that $\lim_{n \to {\infty}}(n^2+n)^{\frac{1}{n}}$...now if we say the limit is equal to y we have $y=\lim_{n \to {\infty}}(n^2+n)^{\frac{1}{n}}$...so then using the everuseful rules of logarithims we have that $ln(y)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln(n^2+n)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln(n(n+1))=\lim_{n \to {\infty}}\frac{1}{n}\cdot[ln(n)+ln(n+1)]$ $=\lim_{n \to {\infty}}\frac{ln(n)}{n}+\frac{ln(n+1)}{n}$...now using lhopitals rule we get $ln(y)=\lim_{n \to {\infty}}\frac{\frac{1}{n}}{1}+\lim_{n \to {\infty}}\frac{\frac{1}{n+1}}{1}=\lim_{n \to {\infty}}\frac{1}{n}+\lim_{n \to {\infty}}\frac{1}{n+1}=0$...so we have $ln(y)=0$ and since y was the original limit we want it by itself..so we have $y=e^0=1$...so the limit converges to the value of e...just so you know I showed you the long way instead of seperating you could have just used l'hopitals rule when you still had $ln(n^2+n)$...I thought I would just take the long road

7. Stop using l'Hôpital's rule !! >_<

It annihilates other methods of calculating limits !

$\lim_{n \to {\infty}}\frac{1}{n}\cdot[ln(n)+ln(n+1)]$

This can be solved by comparing the increasings...

Wikipedia i'm sorry, i can't find any equivalent in english as i don't know the name... It's in "énoncé des résultats..."

ok this is fairly simple
Sure, l'Hôpital's rule does it all. But also stop thinking that because it's simple for you that it'll be simple for everybody.

By the way, i think we're rather talking about series than sequences, as it's $\left\{a_n\right\}$ we're studying.

8. ## Yes

Originally Posted by Moo
Stop using l'Hôpital's rule >_<

It annihilates other methods of calculating limits !

$\lim_{n \to {\infty}}\frac{1}{n}\cdot[ln(n)+ln(n+1)]$

This can be solved by comparing the increasings...

Wikipedia i'm sorry, i can't find any equivalent in english as i don't know the name... It's in "énoncé des résultats..."
I know it can but for the general kid that just wants help understanding the basics of limits...then comparison of increasing would just confuse them generally...I am trying to teach them what they need to pontificate upon advanced mathematical techniques they may or may not understand. Is that ok with you?

9. You know, i've seen previous questions from matty888 and these were not so basic. So i don't think it's a "kid"...
It's certainly not an advance mathematical techniques, these are limits you're supposed to know >_<

10. ## Well

Originally Posted by Moo
You know, i've seen previous questions from matty888 and these were not so basic. So i don't think it's a "kid"...
It's certainly not an advance mathematical techniques, these are limits you're supposed to know >_<
I have never seen any of his questions...so I do what I must and assume the least...I dont want to help by overshooting and give them something too advance that confuses them...I have done that before...