Steply

Does Σ (-1)^(n+1) 3/(n^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⁺¹3/(n² + n) for n from 1 to infinity.

Σn = 1∞(−1)n+13n2 + n =

1 The size

Write the first terms: the signs take turns.

terms = 1.5, −0.5, 0.25, −0.15, ...

Take the size of each term, without the sign: b n = 3/(n² + n).

bn = 3n2 + n

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?

n + 1 is bigger than n, so the bottom gets bigger: each size is smaller than the one before.

bn + 1 < bn

3 Going to 0?

Divide top and bottom by n to the 2.

bn = 3n21 + 1n

Every part with n at the bottom goes to 0.

So the sizes go to 0.

as n → ∞ = 01 + 0 = 0

4 Decide

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

= converges

5 Absolutely or conditionally?

Now drop the signs.

For big n, 3/(n² + n) is like 1/n to the 2: that converges.

Σn = 1∞3n2 + n : ≈ Σ 1n2 , p = 2

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 1.1587, of 101 terms about 1.159.

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

It settles.

Right!

S100 ≈ 1.1587, S101 ≈ 1.159

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)n34n2 + n
Show the answer
converges absolutely

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