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

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  • $\begingroup$ 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. $\endgroup$ – Piotr Wasilewicz Dec 15 '16 at 21:35
  • $\begingroup$ It is equivalent and equivalent sequences (or series) have the same behaviour (and the same limit, for sequences, when one of them has a limit). It's a standard result in asymptotic analysis. $\endgroup$ – Bernard Dec 15 '16 at 21:50

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