sequences

**kastamonu** · Feb 23rd 2013, 11:51 AM

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.

**emakarov** · Feb 23rd 2013, 12:10 PM

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

Secondly, this is not discrete mathematics, but calculus.

**kastamonu** · Feb 23rd 2013, 12:22 PM

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

**emakarov** · Feb 23rd 2013, 12:27 PM

Originally Posted by kastamonu

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

Sorry, this won't work either.

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

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

**Plato** · Feb 23rd 2013, 01:20 PM

Originally Posted by kastamonu

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

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

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

**kastamonu** · Feb 23rd 2013, 01:23 PM

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

**Plato** · Feb 23rd 2013, 01:47 PM

Originally Posted by kastamonu

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

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

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

Or $\left( {1 + \frac{a}{{bn + c}}} \right)^{dn} = \left( {\left( {1 + \frac{1}{m}} \right)^m } \right)^{\frac{{ad}}{b}} \left( {1 + \frac{1}{m}} \right)^{ - \frac{c}{b}}$ .

**kastamonu** · Feb 23rd 2013, 01:58 PM

Many Thanks.

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