Find the limit of (1+(4/3n+1))^6n.

I found it as e^6. [(1+(4/3/n+1)/(3))^6]^n and by the equation (1+1/x)^x=e,it is e^6.

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- Feb 23rd 2013, 11:51 AMkastamonusequences
Find the limit of (1+(4/3n+1))^6n.

I found it as e^6. [(1+(4/3/n+1)/(3))^6]^n and by the equation (1+1/x)^x=e,it is e^6. - Feb 23rd 2013, 12:10 PMemakarovRe: sequences
$\displaystyle (1+4/3n+1)^{6n} = \left(2+\frac{4}{3n}\right)^{6n} > 2^{6n}\to\infty$ as $\displaystyle n\to\infty$.

Secondly, this is not discrete mathematics, but calculus. - Feb 23rd 2013, 12:22 PMkastamonuRe: sequences
answer is e^8 but I can't get it.

- Feb 23rd 2013, 12:27 PMemakarovRe: sequences
- Feb 23rd 2013, 01:20 PMPlatoRe: sequences
- Feb 23rd 2013, 01:23 PMkastamonuRe: sequences
Yes Plato. Many Thanks. But is there a way to derive this formula?

- Feb 23rd 2013, 01:47 PMPlatoRe: sequences

Write it as $\displaystyle \left( {1 + \frac{{\frac{a}{b}}}{{n + \frac{c}{b}}}} \right)^{dn} $.

Let $\displaystyle \frac{1}{m} = \frac{{\frac{a}{b}}}{{n + \frac{c}{b}}}$ so $\displaystyle n=\tfrac{am}{b}-\tfrac{c}{b}$.

Or $\displaystyle \left( {1 + \frac{a}{{bn + c}}} \right)^{dn} = \left( {\left( {1 + \frac{1}{m}} \right)^m } \right)^{\frac{{ad}}{b}} \left( {1 + \frac{1}{m}} \right)^{ - \frac{c}{b}} $. - Feb 23rd 2013, 01:58 PMkastamonuRe: sequences
Many Thanks.