Series converges or diverges

Question

∞

∑ ((3n-2)/(5n+1))^n

n=1

Answer 1

Whenever the whole term is raised to the n^th power, reach for the Root Test first. That exponent is a signal, and it makes this problem almost disappear.

Apply the Root Test

The test says: compute L = lim_n→∞ ⁿ√|a_n|. If L < 1 the series converges absolutely, if L > 1 it diverges, and if L = 1 the test tells you nothing.

Here a_n = ((3n − 2)/(5n + 1))ⁿ, so taking the n^th root cancels the exponent completely:

ⁿ√|a_n| = (3n − 2)/(5n + 1)

That's the whole trick — you're left with an ordinary rational limit. Divide numerator and denominator by n:

lim_n→∞ (3 − 2/n)/(5 + 1/n) = 3/5

Since L = 3/5 < 1, the series converges absolutely.

What's really going on

It helps to see why 3/5 settles the question. For large n the fraction (3n − 2)/(5n + 1) is hovering near 3/5, so the terms behave roughly like (3/5)ⁿ — a geometric series with ratio less than 1. The terms shrink geometrically, and geometric decay is fast enough to sum to something finite.

That also tells you where the cutoff sits. If the leading coefficients had been 5n and 3n instead, the limit would be 5/3 > 1, the terms would grow without bound, and the series would diverge. And if numerator and denominator had matching leading coefficients — say (3n − 2)/(3n + 1) — the limit would be exactly 1 and the Root Test would be useless; you'd need a different approach.

A caution on the limit

The convergence here is slower than it looks. At n = 10 the n^th root is only about 0.549, and at n = 100 it's about 0.595 — still short of 0.6. The limit is what matters, not how fast you get there, but this is a good reminder not to judge a limit by plugging in a few values.

For reference, the sum settles near 0.5129.

The habit worth building: when you see a whole expression raised to the n^th power, try the Root Test before anything else. Reaching for the Ratio Test here would leave you simplifying a messy quotient of two n^th powers for no reason.