Convergence and Divergence with Series

[shamieh]

Determine whether the series is convergent or divergent.

\(\displaystyle \sum^{\infty}_{n = 1} \frac{n - 1}{3n - 1}\)

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

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

Determine whether the series is convergent or divergent.

\(\displaystyle \sum^{\infty}_{n = 1} \frac{1 + 2^n}{3^n}\)

I split this up into two summations

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

 

[alexmahone]

Re: Convergene and Divergence with Series

For \(\displaystyle \sum^{\infty}_{n = 1} \frac{n - 1}{3n - 1}\),

\(\displaystyle \lim a_n=\frac{1}{3}\neq 0\)

So, the series diverges by the divergence test.

\(\displaystyle \sum^{\infty}_{n = 1} \frac{1 + 2^n}{3^n}\) converges to \(\displaystyle \frac52\).