Convergence of series and limit

Question

Let $y_{k\in N}$ $\subset R^2 $.

How do I find out whether this complicated series converges or not and find its limit?

Answer 1

We can see that $y_k^{(1)}$ is the Taylor series of $e^{2-\pi i}$ with the first term missing therefore $y_k^{(1)} = e^{2-\pi i} -1 = e^2e^{-\pi i} - 1 = -e^2 -1$.

For $y_k^{(2)}$, we have $$\frac{k^2+1}{k^2-2} = 1 + \frac{3}{k^2-2}$$ Let $x = k^2-2$. If $k \rightarrow \infty$ then $x = k^2-2 \rightarrow \infty$, therefore $$\lim_{k\rightarrow \infty}y_k^{(2)} = \lim_{x \rightarrow \infty} (1 + \frac{3}{x})^x = e^3$$ Hence $y_k \rightarrow (-e^2-1,e^3)$.

Answer 2

Hints: 1) $y_k$ converges if and only if each coordinate converges.

2) The first coordinate looks a lot like partial sums of the Taylor series of a familiar function

3) For the second coordinate, $$\lim_{k\to\infty} \left(\frac{k^2+1}{k^2-2}\right)^{k^2-2} = \lim_{k\to\infty} \left(1+\frac{3}{k^2-2}\right)^{k^2-2} = \lim_{t\to\infty} \left(1+\frac{3}{t}\right)^t.$$