sum_(n=1)^(oo)(1)/(4+e^(-n))

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Given Series Analysis: We are given an infinite series $ \sum_{n=1}^{\infty} \frac{1}{4+e^{-n}} $. To find the sum of this series, we need to determine if it is a geometric series or if it can be simplified to a known series sum formula. However, this series is not a geometric series, and there is no simple formula for its sum. Therefore, we need to consider other methods to find the sum, such as convergence tests or approximations.Comparison Test: Since the series is not a standard type for which we have a formula, we can check if the series converges using the comparison test. We compare our series to a simpler series that we know converges or diverges. A good comparison might be the series $ \sum_{n=1}^{\infty} \frac{1}{4+n} $, since $ e^{-n} $ approaches 0 as $ n $ goes to infinity, and thus $ 4+e^{-n} $ approaches $ 4+n $.Comparison with Divergent Series: The series $ \sum_{n=1}^{\infty} \frac{1}{4+n} $ is a p-series with $ p = 1 $, which is known to diverge. However, since $ 4+e^{-n} $ is always greater than $ 4+n $, we have $ \frac{1}{4+e^{-n}} < \frac{1}{4+n} $. This means that our series is bounded above by a divergent series, which does not allow us to conclude convergence. We need to find another series to compare it to or another method to determine convergence.Limit Comparison Test: Let's compare our series to the series $ \sum_{n=1}^{\infty} \frac{1}{n} $, which is also a p-series with $ p = 1 $ and is known to diverge. Since $ e^{-n} $ is positive for all $ n $, we have $ 4+e^{-n} > 4 $, and therefore $ \frac{1}{4+e^{-n}} < \frac{1}{4} $. This comparison is not useful because it only tells us that the terms of our series are less than $ \frac{1}{4} $, which does not help in determining convergence.Limit of Ratio of nth Terms: We can use the limit comparison test with the series $ \sum_{n=1}^{\infty} \frac{1}{n} $. We take the limit of the ratio of the nth term of our series to the nth term of the $ \frac{1}{n} $ series as $ n $ approaches infinity. If the limit is finite and positive, then both series either converge or diverge together.Inconclusive Limit Comparison Test: Calculate the limit of the ratio of the nth terms: $$ \lim_{n \to \infty} \frac{\frac{1}{4+e^{-n}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{4+e^{-n}} $$ As $ n $ approaches infinity, $ e^{-n} $ approaches 0, so the limit simplifies to: $$ \lim_{n \to \infty} \frac{n}{4} = \infty $$ This means that the limit comparison test is inconclusive because the limit is not finite.Ratio Test: Since the limit comparison test was inconclusive, we can try the ratio test to determine the convergence of the series. The ratio test states that for a series $ \sum a_n $, if the limit $ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $ exists and $ L < 1 $, the series converges; if $ L > 1 $, the series diverges; and if $ L = 1 $, the test is inconclusive.Application of Ratio Test: Apply the ratio test to our series: $$ L = \lim_{n \to \infty} \left| \frac{\frac{1}{4+e^{-(n+1)}}}{\frac{1}{4+e^{-n}}} \right| = \lim_{n \to \infty} \left| \frac{4+e^{-n}}{4+e^{-(n+1)}} \right| $$ Simplify the expression inside the limit: $$ L = \lim_{n \to \infty} \frac{4e^{n+1}+1}{4e^n+1} $$As $ n $ approaches infinity, both the numerator and the denominator grow without bound, but the exponential terms dominate, so we can divide both the numerator and the denominator by $ e^n $ to get: $$ L = \lim_{n \to \infty} \frac{4e+e^{-n}}{4+e^{-n}} $$ Now, as $ n $ approaches infinity, $ e^{-n} $ approaches 0, so the limit simplifies to: $$ L = \frac{4e}{4} = e $$ Since $ e > 1 $, the ratio test tells us that the series diverges.

$\sum_{n=1}^{\infty}\frac{1}{4+e^{-n}}$

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