Verified answer

Answer:

[tex]\boxed{ \sf{ \:\bf \:\displaystyle\lim_{n \to \infty }\bf \: \frac{n + {n}^{2} + {n}^{3} + ...... {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + ..... {n}^{n} } = \: \dfrac{e - 1}{e} \: }} \\ \\ [/tex]

Step-by-step explanation:

Given expression is

[tex]\sf \: \displaystyle\lim_{n \to \infty }\sf \: \frac{n + {n}^{2} + {n}^{3} + ...... {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + ..... {n}^{n} } \\ \\ [/tex]

Since,

[tex]\sf \: n, {n}^{2}, {n}^{3},...,{n}^{n} \: forms \: a \: geometric \: sequence \\ \\ [/tex]

So, using sum of n terms of GP series having first term n and common ratio n and number of terms n is

[tex]\sf \: n + {n}^{2} + {n}^{3} + ... + {n}^{n} \: = \dfrac{n( {n}^{n} - 1)}{n - 1} \\ \\ [/tex]

So, given expression can be rewritten as

[tex]\sf \: = \displaystyle\lim_{n \to \infty }\sf \: \frac{\dfrac{n( {n}^{n} - 1) }{n - 1} }{ {1}^{n} + {2}^{n} + {3}^{n} + ..... {n}^{n} } \\ \\ [/tex]

[tex]\sf \: = \displaystyle\lim_{n \to \infty }\sf \:\dfrac{n }{n - 1} \times \displaystyle\lim_{n \to \infty }\sf \: \frac{ {n}^{n} - 1}{ {1}^{n} + {2}^{n} + {3}^{n} + ..... {n}^{n} } \\ \\ [/tex]

[tex]\sf \: = 1\times \displaystyle\lim_{n \to \infty }\sf \: \frac{ {n}^{n} - 1}{ {n}^{n}\bigg( {\bigg(\dfrac{1}{n} \bigg) }^{n} + {\bigg(\dfrac{2}{n} \bigg) }^{n} + {\bigg(\dfrac{3}{n} \bigg) }^{n} - - + {\bigg(\dfrac{n}{n} \bigg) }^{n} \bigg) } \\ \\ [/tex]

[tex]\sf \: = \displaystyle\lim_{n \to \infty }\sf \: \frac{1 - \dfrac{1}{ {n}^{n} } }{ {\bigg(\dfrac{1}{n} \bigg) }^{n} + {\bigg(\dfrac{2}{n} \bigg) }^{n} + {\bigg(\dfrac{3}{n} \bigg) }^{n} - - + {\bigg(\dfrac{n}{n} \bigg) }^{n} } \\ \\ [/tex]

[tex]\sf \: = \displaystyle\lim_{n \to \infty }\sf \: \frac{1 - 0 }{ {\bigg(\dfrac{n}{n} \bigg) }^{n} + {\bigg(\dfrac{n - 1}{n} \bigg) }^{n} + {\bigg(\dfrac{n - 2}{n} \bigg) }^{n} - - + {\bigg(\dfrac{1}{n} \bigg) }^{n} } \\ \\ [/tex]

[tex]\sf \: = \displaystyle\lim_{n \to \infty }\sf \: \frac{1 }{ 1+ {\bigg(1 - \dfrac{1}{n} \bigg) }^{n} + {\bigg(1 - \dfrac{2}{n} \bigg) }^{n} - - - - } \\ \\ [/tex]

We know,

[tex]\boxed{ \sf{ \:\displaystyle\lim_{x \to \infty }\sf \:{\bigg(1 + \dfrac{k}{x} \bigg) }^{x} \: = \: {e}^{k} }} \\ \\ [/tex]

So, using this result, we get

[tex]\sf \: = \displaystyle\lim_{n \to \infty }\sf \: \frac{1 }{ 1+ {e}^{ - 1} + {e}^{ - 2} + - - - - } \\ \\ [/tex]

Now,

[tex]\sf \: 1+ {e}^{ - 1} + {e}^{ - 2} + - - - \: forms \: infinite \: geometric \: sequence \\ \\ [/tex]

Here,

[tex]\sf \: a = 1 \\ \\ [/tex]

[tex]\sf \: r = \frac{1}{e} < 1 \\ \\ [/tex]

So, using sum of infinite GP series, we get

[tex]\sf \: 1+ {e}^{ - 1} + {e}^{ - 2} + - - - = \dfrac{1}{1 - \dfrac{1}{e} } \\ \\ [/tex]

[tex]\sf \: 1+ {e}^{ - 1} + {e}^{ - 2} + - - - = \dfrac{1}{\dfrac{e - 1}{e} } \\ \\ [/tex]

[tex]\sf \: 1+ {e}^{ - 1} + {e}^{ - 2} + - - - = \dfrac{e}{e - 1} \\ \\ [/tex]

So, using this, the above expression can be rewritten as

[tex]\sf \: = \: \dfrac{1}{\dfrac{e}{e - 1}} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{e - 1}{e} \\ \\ [/tex]

Hence,

[tex]\boxed{ \sf{ \:\bf \:\displaystyle\lim_{n \to \infty }\bf \: \frac{n + {n}^{2} + {n}^{3} + ...... {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + ..... {n}^{n} } = \: \dfrac{e - 1}{e} \: }} \\ \\ [/tex]