# How can I prove that the value of this series is smaller than a constant c < 1

I am stuck in a proof in graph theory (random graphs) because in the last part, I need to prove the following:

$$\sum_{k=3}^n \frac{1}{4^k e^{2k}} < c$$

for some constant $c < 1$ and any $n \in \mathbb{N}$. I know how to prove the convergence of this series using standard convergence tests, but how do I ensure that this series will actually have the desired value (say, less than one half)?

Thank you very much for your time (and thank you more than much for any helpful hint!)

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Hint: For $0<q<1$ we have $$\sum_{k=3}^n q^k<\sum_{k=3}^\infty q^k=q^3\cdot \sum_{k=0}^\infty q^k=\frac{q^3}{1-q}.$$

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I hope this answers your question: $\frac{1}{4^k e^{2k}} < \frac{1}{4^k}$ for $k \ge 3$.

So $$\sum^n_{k=3}\frac{1}{4^k e^{2k}} < \sum^\infty_{k=3}\frac{1}{4^k} = \frac{1}{48}$$

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From Hagen von Eitzen's answer, set $q = 1/(4 e^2) \approx 0.0338$, so the sum = $q^3/(1-q) \approx 0.0004$.
Note that for all $k\in\mathbb{N}$ we have $4^k\leq e^{2k}$ implies $\frac{1}{e^{2k}}\leq \frac{1}{4^k}$ and $\frac{1}{e^{2k}}\frac{1}{4^{2k}}\leq 1\frac{1}{4^{2k}}$. Then $$\sum_{k=3}^n \frac{1}{4^k e^{2k}}\leq \sum_{k=3}^n \frac{1}{4^{2k}}< \sum_{k=0}^n \frac{1}{(4^{2})^{k}}$$ By somation formula of geometric serie we have: $\sum_{k=0}^n \frac{1}{4^k}=\frac{1-(1/16)^n}{1-1/16}<\frac{1}{1-1/16} < 1$