Thread: sequences

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.

$\displaystyle (1+4/3n+1)^{6n} = \left(2+\frac{4}{3n}\right)^{6n} > 2^{6n}\to\infty$ as $\displaystyle n\to\infty$.

Secondly, this is not discrete mathematics, but calculus.

answer is e^8 but I can't get it.

Originally Posted by kastamonu

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

Sorry, this won't work either.

$\displaystyle (1+(4/3n+1))^{6n} = \left(1+\left(\frac{4}{3n}+1\right)\right)^{6n} = \left(2+\frac{4}{3n}\right)^{6n} > 2^{6n}\to\infty$ as $\displaystyle n\to\infty$.

And there is a question whether 4/3n means (4/3)n or 4/(3n).

Originally Posted by kastamonu

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

I an sure you mean $\displaystyle \left( {1 + \frac{4}{{3n + 1}}} \right)^{6n} $


If $\displaystyle a\cdot b\ne 0$ then $\displaystyle \left( {1 + \frac{a}{{bn + c}}} \right)^{dn} \to e^{\frac{{ad}}{b}} $.

Yes Plato. Many Thanks. But is there a way to derive this formula?

Originally Posted by kastamonu

Yes Plato. Many Thanks. But is there a way to derive this formula?

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}} $.

Many Thanks.