# How to solve these two limits?

I need to solve these limits of a sequences usign squeeze theorem:

1. $$\lim_{n \to \infty} \left( \frac{e-1}{\pi+2} + \frac{e^2-2^{-1}}{\pi^2+2^2} + \cdot\cdot\cdot + \frac{e^n-2^{1-n}}{\pi^n+2^n} \right)$$

I know that

$$\lim_{n \to \infty} \frac{e^n-2^{1-n}}{\pi^n+2^n} = 0$$ and how to compute this value.

1. $$\lim_{n \to \infty} \left( \frac{e}{n^2+\pi} + \frac{2e}{n^2+2\pi} + \cdot\cdot\cdot + \frac{ne}{n^2+n\pi} \right)$$
• Why is there a square root only in the second term of 2? – Bernard Dec 15 '16 at 18:08
• can you calculate these sums? – Dr. Sonnhard Graubner Dec 15 '16 at 18:13
• @Bernard My mistake. Sorry. – Piotr Wasilewicz Dec 15 '16 at 20:25

For convergence of $1$: as this is a series with positive terms you can use equivalents:
$\mathrm e^n-2^{1-n}\sim_\infty\mathrm e^n$, $\;\pi^n+2^n\sim_\infty \pi^n$, hence $\;\dfrac{\mathrm e^n-2^{1-n}}{\pi^n+2^n}\sim_\infty\Bigl(\dfrac{\mathrm e}{\pi}\Bigr)^n$, and the latter converges since $0\le\mathrm e/\pi <1$.
$2$ is divergent since $\;\dfrac{n\mkern 1mu\mathrm e}{n^2+n\pi}\sim_\infty\dfrac{n\mkern 1mu\mathrm e}{n^2}=\dfrac{\mathrm e}{n}$, and the harmonic series is divergent.
• I don't understand the second one. You choose fraction $\frac{ne}{n^2}$ which is bigger that $\frac{ne}{n^2+n\pi}$ and if bigger is divergent it not implies that smaller also is. In first one I need to calculate exact value. – Piotr Wasilewicz Dec 15 '16 at 21:35