Steply

Does Σ (-1)^(n+1)/2^n converge

Step by step, with the reason for every step · Alternating series · Calculus 2

Step by step

The question

Question: decide if the series converges: the sum of (−1)ⁿ⁺¹ over 2ⁿ for n from 1 to infinity.

Σn = 1∞(−1)n+12n =

1 The size

Write the first terms: the signs take turns.

terms = 0.5, −0.25, 0.125, −0.062, ...

Take the size of each term, without the sign: b n = 1/2ⁿ.

bn = 12n

The alternating series test: if the sizes get smaller and smaller and go to 0, the series converges.

alternating series test:
if bn gets smaller and bn → 0,
the series converges

2 Getting smaller?

Each term is the one before times 1/2: smaller.

bn + 1 < bn

3 Going to 0?

A fraction less than 1, to bigger and bigger powers, goes to 0.

as n → ∞ = 12n → 0

4 Decide

Both checks pass: the series converges, by the alternating series test.

= converges

5 Absolutely or conditionally?

Now drop the signs: a geometric series with r = 1/2.

It converges.

Σn = 1∞12n : geometric, r = 1 over 2 < 1

The series converges even without its signs.

That is called converging absolutely.

Σ bₙ converges too: absolutely

So the series converges absolutely.

= converges absolutely

Careful!

Absolutely means it would converge even with all the signs plus.

Careful: absolutely: it converges even without the signs

6 Check

Let's check with a calculator: the sum of 100 terms is about 0.3333, of 101 terms about 0.3333.

They jump less and less, and the sum is between them.

It settles.

Right!

S100 ≈ 0.3333, S101 ≈ 0.3333

Let's look back.

First: the size.

Then: getting smaller?.

Then: going to 0?.

Then: decide.

Then: absolutely or conditionally?.

Last: check.

Answer
= converges absolutely

Your turn

Try this one yourself, then check:

Σn = 1∞(−1)n+133n
Show the answer
converges absolutely

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