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