I am trying to find the limit of the sequence $$ \left(\frac{3-4n}{1+n}\right)\left(1+\frac1n\right)^n $$ I am aware that if one sequence converges and another sequence converges then the multiplication of two sequences also converge. The limit of the first sequence is $-4$. However I do not know how to calculate the limit of the second sequence.
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What is n tending to? – Rohinb97 Aug 20 '13 at 16:10
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@Robinb97 n is tending towards infinity. Apologies, I forgot to put that up – Anona anon Aug 20 '13 at 16:11
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1It can be done with Maple. See the screen from RapidShare. – user64494 Aug 20 '13 at 17:21
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Here is one approach $$ \left( 1+\frac{1}{n}\right)^{n}=e^{ n \ln(1+\frac{1}{n}) } = e^{ n (\frac{1}{n}-\frac{1}{2 n^2}+\dots) } = e^{1-\frac{1}{2n}+\dots}\underset{\infty}{\longrightarrow} e $$
robjohn
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Mhenni Benghorbal
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@ MhenniBenghorbal Thank you for your help. So would the final limit of the entire product be -4e? – Anona anon Aug 20 '13 at 16:10
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This is the constant known as $e$. In this answer, it is shown to be equal to $$ \sum_{n=0}^\infty\frac1{n!}\tag{1} $$ In this answer, Bernoulli's Inequality is used to show that $$ \left(1+\frac1n\right)^n\tag{2} $$ is an increasing sequence and that $$ \left(1+\frac1n\right)^{n+1}\tag{3} $$ is a decreasing sequence. Thus, $e$ is greater than any term in $(2)$ and less than any term in $(3)$.
$n=1$ in $(2)$ shows that $e\ge2$ and $n=5$ in $(3)$ shows that $e\lt3$.
