# Limit of series

1. Feb 14, 2006

### teng125

for infinity sum n=1 (2n+1) / (5n+1)

how to find the converges or diverges??
is it suitable to use basic comparison test??pls show me

2. Feb 14, 2006

### TD

If we split the fraction, we get

$$\frac{{2n + 1}}{{5n + 1}} = \frac{2}{5} + \frac{3}{{25n + 5}}$$

As you can see, the second term will go to 0 if n tends to infinity but there's a remaining term 2/5. In other words: the limit of the associated sequence isn't 0 (for n going to infinity) and this is a necessary (though not sufficient) condition for the convergence of the series.

3. Feb 14, 2006

### HallsofIvy

Or: divide both numerator and denominator by n:
$$\frac{2n+1}{5n+1}= \frac{2+\frac{1}{n}}{5+\frac{1}{n}}$$

Now what does 1/n go to as n goes to infinity?

4. Feb 14, 2006

### teng125

but the question ask for diverges or converges??and the answer is diverges

5. Feb 14, 2006

### TD

When a series doesn't converge, it diverges.

6. Feb 14, 2006

### teng125

is it necessary to use any rules such as the one i mentioned??or just to perform the steps showed above??

7. Feb 14, 2006

### TD

I don't think those are necessary, if the limit of the (positive) sequence doesn't go to 0 (which is easy to check without using those other tests), then the (positive) series doesn't converge.

8. Feb 14, 2006

### Valhalla

teng the n-th term test for divergence is an easy check for this one.