Evaluate limx-1n-e1-nn-e2-nn-e3-nn-en-1-nn-

lim_x→∞1n+e^1/nn+e^2/nn+e^3/nn+...+e^n 1/nn =lim_x→∞1n[1+e^1/n+e^2/n+e^3/n+...+e^n 1/n] =lim_x→∞1n+lim_x→∞[e^1/n+e^2/n+e^3/n+...+e^n 1/n] =0 ...

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Byju's Answer

Standard XII

Mathematics

Proof by mathematical induction

Evaluate li...

Question

Evaluate $lim x \to \infty \frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} + \frac{e^{\frac{2}{n}}}{n} + \frac{e^{\frac{3}{n}}}{n} + . . . + \frac{e^{\frac{n - 1}{n}}}{n}$ .

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Solution

$lim x \to \infty \frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} + \frac{e^{\frac{2}{n}}}{n} + \frac{e^{\frac{3}{n}}}{n} + . . . + \frac{e^{\frac{n - 1}{n}}}{n}$

$= lim x \to \infty \frac{1}{n} [1 + e^{\frac{1}{n}} + e^{\frac{2}{n}} + e^{\frac{3}{n}} + . . . + e^{\frac{n - 1}{n}}]$

$= lim x \to \infty \frac{1}{n} + lim x \to \infty [e^{\frac{1}{n}} + e^{\frac{2}{n}} + e^{\frac{3}{n}} + . . . + e^{\frac{n - 1}{n}}]$

$= 0 + lim x \to \infty \frac{1}{n} n \sum r = 1 e^{\frac{r}{n}}$ by using the definition of integral we have

$= \int_{0}^{1} e^{x} d x = {[e^{x}]}_{0}^{x} = e^{1} - e^{0} = e - 1$

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Similar questions

{lim}_{n \to \infty} (\frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} + \frac{e^{\frac{2}{n}}}{n} + . . . + \frac{e^{\frac{n - 1}{n}}}{n})

equals

Q. The value of

lim n \to \infty [\frac{1}{n} + \frac{e^{1 / n}}{n} + \frac{e^{2 / n}}{n} + . . . + \frac{e^{(n - 1) / n}}{n}]

lim n \to \infty \frac{1}{n} ⎛ ⎜ ⎝ 1 + e^{1 / n} + e^{2 / n} + . . . + e^{\frac{n - 1}{n}} ⎞ ⎟ ⎠

is equal to

lim n \to \infty \frac{1}{n} [e^{1 / n} + e^{2 / n} + \dots . . + e] = ?

If $E_{1}, E_{2}, E_{3} \dots E_{n}$ are n events of a sample space S and if $E_{1} \cup E_{2} \cup E_{3} \cup \dots E_{n} = S$ , then $E_{1}, E_{2} \dots E_{n}$ are called mutually exclusive and exhaustive events.

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Evaluate limx→∞1n+e1nn+e2nn+e3nn+...+en−1nn.

limx→∞1n+e1nn+e2nn+e3nn+...+en−1nn=limx→∞1n[1+e1n+e2n+e3n+...+en−1n]=limx→∞1n+limx→∞[e1n+e2n+e3n+...+en−1n]=0+limx→∞1nn∑r=1ern by using the definition of integral we have=∫10exdx=[ex]10=e1−e0=e−1

Evaluate $lim x \to \infty \frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} + \frac{e^{\frac{2}{n}}}{n} + \frac{e^{\frac{3}{n}}}{n} + . . . + \frac{e^{\frac{n - 1}{n}}}{n}$ .