MythicalNateidn
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[tex]\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{ n } (1 - e^{\frac{-jt}{n}} )[/tex]
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Chaser27idn

[tex] \large \bold{PROBLEM:}[/tex]

[tex]\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{ n } (1 - e^{\frac{-jt}{n}} )[/tex]

[tex] \large \bold{SOLUTION:}[/tex]

[tex] \bold{look \: at \: \frac{1}{n} \sum_{j=1}^{ n } (1 - e^{\frac{-jt}{n}} )} \\ \bold{e^{\frac{-jt}{n}} =\sum_{k=0}^{\infty} \frac{(-jt/n)^k}{k!}} \\ \bold{1-e^{\frac{-jt}{n}} =-\sum_{k=1}^{\infty} \frac{(-jt/n)^{k}}{k!}} \\ \\ \bold{\begin{align} \sum_{j=1}^{ n } (1 - e^{\frac{-jt}{n}} ) &=\sum_{j=1}^{ n }\sum_{k=1}^{\infty} -\frac{(-jt/n)^{k}}{k!}\\ &=-\sum_{k=1}^{\infty}\sum_{j=1}^{ n } \frac{(-jt/n)^{k}}{k!}\\ &=-\sum_{k=1}^{\infty}\frac{(-t/n)^{k}}{k!}\sum_{j=1}^{ n } j^{k}\\ \end{align}}[/tex]

[tex] \large \bold{Using \: this, \: the \: sum \: is}[/tex]

[tex] \bold{\begin{align} -\sum_{k=1}^{\infty}\frac{(-t/n)^{k}}{k!} (\frac{n^{k+1}}{k+1}+O(n^k)) &=-n \sum_{k=1}^{\infty}\frac{(-t)^{k}}{(k+1)!} +O(\sum_{k=1}^{\infty}\frac{(-t)^{k}}{k!})\\ &=\frac{-n}{-t} \sum_{k=1}^{\infty}\frac{(-t)^{k+1}}{(k+1)!} +O(e^{-t}-1)\\ &=\frac{n}{t} (e^{-t}-1+t) +O(1-e^{-t})\\ \end{align}}[/tex]

[tex] \bold{The \: final \: result \: is \:  1/n1/n  \: times \: this \: or}[/tex]

[tex] \bold{\frac{1}{t} (e^{-t}-1+t) +O((1-e^{-t})/n)} \\ \bold{as \: n \to \infty} \\ \bold{this \: become \: 1-(1-e^{-t})/t}[/tex]

[tex]\purple{\begin{gathered} \gamma \\ \huge \boxed{ \ddot \smile}\end{gathered}} \: \pink{\begin{gathered} \gamma \\ \huge \boxed{ \ddot \smile}\end{gathered}} \: \red{\begin{gathered} \gamma \\ \huge \boxed{ \ddot \smile}\end{gathered}} \: \orange{\begin{gathered} \gamma \\ \huge \boxed{ \ddot \smile}\end{gathered}}[/tex]

EngrJohannidn

[tex] \Large \mathbb{SOLUTION:} [/tex]

[tex] \begin{array}{l} \displaystyle \sf \lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{ n } (1 - e^{\frac{-jt}{n}}) \\ \\ =\small \displaystyle \sf \lim_{n \to \infty} \frac{1}{n} \left(n - \dfrac{e^{\frac{-t}{n}}(e^{-t} - 1)}{e^{\frac{-t}{n}} - 1}\cdot \dfrac{e^{\frac{t}{n}}}{e^{\frac{t}{n}}} \right) \textsf{(By Geometric Series)} \\ \\ = \displaystyle \sf 1 - \lim_{n \to \infty} \dfrac{e^{-t}-1}{n(1 - e^{\frac{t}{n}})} \\ \\ = \displaystyle \sf 1 - (e^{-t}-1) \lim_{n \to \infty} \dfrac{1}{n(1 - e^{\frac{t}{n}})} \\ \\ = \displaystyle \sf 1 - (e^{-t}-1) \lim_{n \to \infty} \dfrac{\frac{1}{n}}{1 - e^{\frac{t}{n}}} \\ \\ \textsf{Let }\sf u = \dfrac{1}{n}. \\ \\ \quad \displaystyle \sf \lim_{n \to \infty} \frac{1}{n} = 0, \textsf{ so }u \to 0. \\ \\ = \displaystyle \sf 1 - (e^{-t}-1) \lim_{u \to 0} \dfrac{u}{1 - e^{tu}} \\ \\ \textsf{Substituting }\sf u =0, \textsf{the limit results to one of} \\ \textsf{the indeterminate forms }\sf \frac{0}{0}. \\ \\ \textsf{Applying l'Hôpital's rule, we get} \\ \\ = \displaystyle \sf 1 - (e^{-t}-1) \lim_{u \to 0} \dfrac{\frac{d}{du} u}{\frac{d}{du}(1 - e^{tu})} \\ \\ = \displaystyle \sf 1 - (e^{-t}-1) \lim_{u \to 0} -\dfrac{1}{te^{tu}} \\ \\ = \displaystyle \sf 1 - (e^{-t}-1)\left(-\dfrac{1}{te^{t(0)}}\right) \\ \\ = \displaystyle \sf 1 - (e^{-t}-1)\left(-\dfrac{1}{t}\right) \\ \\ = \displaystyle \boxed{\sf 1 + \dfrac{e^{-t}-1}{t}} \end{array} [/tex]