# Sequences and series

• January 25th 2009, 03:45 AM
Sequences and series
Determine if the following series is convergent or divergent .

$3-\frac{1}{2n},3-\frac{2}{3n},3-\frac{3}{4n},...$

I know it is a convergent series and it converges to 3 but i am unsure how to show it .

I tried ..

$\lim _{n \rightarrow \infty }{u_n}=3$

I am also not sure how to find $u_n$ ..

Thanks ..
• January 25th 2009, 03:59 AM
Prove It
Quote:

Determine if the following series is convergent or divergent .

$3-\frac{1}{2n},3-\frac{2}{3n},3-\frac{3}{4n},...$

I know it is a convergent series and it converges to 3 but i am unsure how to show it .

I tried ..

$\lim _{n \rightarrow \infty }{u_n}=3$

I am also not sure how to find $u_n$ ..

Thanks ..

First of all, what you've given is not a series, but a sequence.

If you had a sum of all these terms, then you would have a series.

If you want to show that this sequence converges as $n \to \infty$, notice that n is in the denominator. As n gets large, the fraction gets small (tends to 0).

So what would $3 - \frac{a}{n}$ tend to?
• January 25th 2009, 07:56 AM
Opalg
Quote:

Determine if the following series is convergent or divergent .

$3-\frac{1}{2n},3-\frac{2}{3n},3-\frac{3}{4n},...$

I know it is a convergent series and it converges to 3 but i am unsure how to show it .

I tried ..

$\lim _{n \rightarrow \infty }{u_n}=3$

I am also not sure how to find $u_n$ ..

Thanks ..

For a start, the symbol n occurs in the definition of the terms in the sequence. So n is presumably meant to be a fixed constant in this problem, and you should not talk about the n'th term in the sequence. You need to use another letter. Let's refer to the k'th term in the sequence.

If the first few terms in the sequence are $3-\frac{1}{2n},3-\frac{2}{3n},3-\frac{3}{4n},...$, then the k'th term will be $u_k = 3- \frac k{(k+1)n}$, and you need to find the limit $3-\lim_{k\to\infty}\frac k{(k+1)n}$.

Can you take it from there? Don't forget that it is k that goes to infinity, and n stays as a constant.
• January 25th 2009, 08:01 AM