Is there convergence in this sequence?

Question

I have the sequence

$a_{n} = \frac{1}{5n^2 + \cos(n\pi )+1}$ $n\in \mathbb{N}$

It's obvious that it converges to 0 but I have problems to proof it:

Let $\varepsilon$ be optional and choose $N$ as $N\gt?$

Then apply to all $a_{n}$ with $n\ge N$:

$\left | a_{n}-0 \right |$ = $\frac{1}{5n^2 + \cos(n\pi )+1}$

Because of $n\ge N$:

$\frac{1}{5n^2 + \cos(n\pi )+1} \le \frac{1}{5N^2 + \cos(N\pi )+1}$

And now I try to solve for $N$?

$\frac{1}{5N^2 + \cos(N\pi )+1} \gt \varepsilon$

Is that right so far? And after that I might go with a distinction of cases?

Answer 1

As $\mathrm{cos}(n\pi) = (-1)^n$ we get:

$$ 0\leq \frac{1}{5n^2+2} \leq a_n \leq\frac{1}{5n^2}.$$

Answer 2

It is $$0 \le a_n \le\frac{1}{5} \times \frac{1}{n^2}.$$

1) We can use Squeeze Theorem.

2)(a bit fantasy and answer for the related series) We can use Comparison test for related positive series. If series is convergent, then limit of sequence has to be $0$.

Answer 3

Not quite. Use $|a_n| \leq \dfrac{1}{5n^2}$.

Answer 4

The key here is to realize that the cosine part doesn't really matter, since it's only a difference of 1 at the most, so x + cos(x) ≥ x - 1.

5n² + cos(nπ) + 1 ≥ 5n² - 1 + 1 = 5n²
==> 1/(5n² + cos(nπ) + 1) ≤ 1/(5n^2)
By the Archimedean property, we know there exists N ≥ √1/(5ε)
Rearranging, for n≥N: ε ≥ 1/5N^2 ≥ 1/5n^2 ≥ 1/(5n² + cos(nπ) + 1) as desired.