You should know that $\displaystyle {\lim _{n \to \infty }}{\left( {1 + \frac{a}{{n + b}}} \right)^{cn}} = {e^{ac}}$
Well $\displaystyle {\lim _{k \to \infty }}{\left( {\frac{k}{{k + 1}}} \right)^k} = {\lim _{k \to \infty }}{\left( {1 + \frac{{ - 1}}{{k + 1}}} \right)^k}$
Alternatively, represent $\displaystyle \left(\frac{x}{x+1}\right)^x$ as $\displaystyle e^{x\ln(x/(x+1))}$ and then prove that $\displaystyle x\ln(x/(x+1))\to-1$ as $\displaystyle x\to\infty$ using either Taylor series or L'Hopital's rule.
Hello, kaya2345!
We are expected to know this definition:
. . $\displaystyle \lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^n \;=\;e$
Prove that that the following series diverges: .$\displaystyle \sum^{\infty}_{k=1}\left(\frac{k}{k+1}\right)^k$
Theorem: If the n^{th} term does not approach zero, the series diverges.
We have: .$\displaystyle a_n \;=\;\left(\frac{k}{k+1}\right)^k \;=\;\left(\frac{1}{\frac{k+1}{k}}\right)^k \;=\;\left(\frac{1}{1+\frac{1}{k}}\right)^k \;=\;\frac{1}{(1+\frac{1}{k})^k} $
Hence: .$\displaystyle \lim_{k\to\infty}a_n \;=\;\lim_{k\to\infty}\frac{1}{(1+\frac{1}{k})^k} \;=\;\frac{1}{\lim(1+\frac{1}{k})^k} \;=\;\frac{1}{e} \;\ne\;0 $
Therefore, the series diverges.