Steply

Does Σ 3/(4^n + 3) converge

Step by step, with the reason for every step · Comparison tests · Calculus 2

Step by step

The question

Question: decide if the series converges: the sum of 3/(4ⁿ + 3) for n from 1 to infinity.

Σn = 1∞34n + 3 =

1 For big n

For big n, 4 to the n is huge: the + 3 hardly matters.

So the terms are about 3/4ⁿ.

an = 34n + 3 ≈ 34n

2 The known series

The sum of 3/4ⁿ is a geometric series: each term is 1/4 of the one before.

r is less than 1, so it converges.

Σn = 1∞34n: r = 14 < 1, converges

3 Compare

The bottom of aₙ is bigger: it has 3 more.

4n + 3 > 4n

A bigger bottom makes a smaller fraction.

So every term is smaller than 3/4ⁿ.

an < 34n

The direct comparison test: our terms are smaller than terms that add up to a number.

So ours add up to a number too.

smaller than a converging series: converges

4 Decide

So our series converges.

= converges

Careful!

Smaller than a series that diverges tells us nothing.

Compare with one that converges.

Careful: not an < 1n, so it diverges ✗

5 Check

Let's check with a calculator: 100, 1000 and 10000 terms add up to about 0.6467, 0.6467 and 0.6467.

It settles.

Right!

S100 ≈ 0.6467, S1000 ≈ 0.6467, S10000 ≈ 0.6467

Let's look back.

First: for big n.

Then: the known series.

Then: compare.

Then: decide.

Last: check.

Answer
= converges

Your turn

Try this one yourself, then check:

Σn = 1∞14n + 3
Show the answer
converges

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