1. ## Series

Hi, I couldnt do these 3 questions. Anyone can help which method to use please?

I tried divergence test for d. but I got 0 which means divergence test failed. and....

h) I used l'ho and got 1. is that correct?

2. ## Re: Series

The first is a series with $$a_n=-\frac{1}{n(n+1)}$$ here $|a_n|<1/(n+1)^2$ and since $\sum_{i=2}^{\infty}1/n^2$ converges the series here converges absolutely and so converges by the comparison test.

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3. ## Re: Series

Originally Posted by csc

Hi, I couldnt do these 3 questions. Anyone can help which method to use please?

I tried divergence test for d. but I got 0 which means divergence test failed. and....

h) I used l'ho and got 1. is that correct?
For the second two you should note that $\lim_{n \to \infty} a_n =1$ so they do not converge.

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4. ## Re: Series

Originally Posted by zzephod
For the second two you should note that $\lim_{n \to \infty} a_n =1$ so they do not converge.

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$\displaystyle\[{\lim _{x \to \infty }}\frac{{\sin \left( {{n^{ - 1}}} \right)}}{{{n^{ - 1}}}} = 1$