Convergence and Divergence with Series

[shamieh]

Determine whether the series is convergent or divergent.

$$\sum^{\infty}_{n = 1} \frac{n - 1}{3n - 1}$$

I ended up with $$\frac{1}{3} * 1 = \frac{1}{3}$$ , which is 0.333 ... so wouldn't that mean that $$r < 1$$? Also wouldn't that mean that it is convergent since $$r < 1$$ ?

I don't understand why this is actually divergent?Also,

Determine whether the series is convergent or divergent.

$$\sum^{\infty}_{n = 1} \frac{1 + 2^n}{3^n}$$

I split this up into two summations

Ended up with $$\frac{5}{2}$$ and r < 1, so this converges at $$\frac{1}{3}$$ while approaching infinity right?

 

[alexmahone]

Re: Convergene and Divergence with Series

For $$\sum^{\infty}_{n = 1} \frac{n - 1}{3n - 1}$$,

[math]\lim a_n=\frac{1}{3}\neq 0[/math]

So, the series diverges by the divergence test.

$$\sum^{\infty}_{n = 1} \frac{1 + 2^n}{3^n}$$ converges to [math]\frac52[/math].