# MAth problem (picture)

Math problem (picture) (second problem added)

Can anyone explain this simplification?

http://img379.imageshack.us/img379/1786/maproblem1bm.jpg [Broken]

Kindly Paul-MArtin

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TD
Homework Helper
You probably know that dividing by a fraction is the same as multiplying it with the inverse. So that (a/b)/(c/d) = (a/b)*(d/c). Applying that here gives

$$\frac{{\frac{{3^{k + 1} }}{{\left( {k + 1} \right)!}}}}{{\frac{{3^k }}{{k!}}}} = \frac{{3^{k + 1} }}{{\left( {k + 1} \right)!}}\frac{{k!}}{{3^k }} = \frac{{3 \cdot 3^k \cdot k!}}{{3^k \left( {k + 1} \right) \cdot k!}} = \frac{3}{{\left( {k + 1} \right)}}$$

Very good, thank you!,

convincing myself!
K=1
1*2=(1+1)*1 ok

k=2
1*2*3=(3*1*2)

k=3
1*2*3*4=(4*1*2*3)

Now i se it's obvious stupid me :(...

I wonder can anyone simplifi this problem? what am i missing?

http://img55.imageshack.us/img55/1434/maproblem22pi.jpg [Broken]

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What? That problem doesn't make any sense. Are you sure you've written it down correctly?

HallsofIvy
Homework Helper
1. Your picture says $$lim_{x->\infty}$$ but there is no x in the formula. Did you mean k instead of x?

2. If the limit you show equals any y, then, by definition, it is not divergent!!

I think you are referring to the "ratio" test for infinite series:
If the limit $$lim_{k->\infty}\frac{a_{n+1}}{a_n}= y$$
and y< 1 then the infinite series $$\Sigma_{k=1}^{\infty}a_n$$
converges, if y>1 then it diverges.

The general proof of that is given in any calculus text. It basically uses y to compare the series to a geometric series.

If, eventually, $$\frac{a_{k+1}}{a_k}< y$$ then we can write ak+1< yak< y2ak-1< ....< yka1 so that the series in dominated by the geometric series $$\Sigma_{k=1}^{\infty}a_1y^k$$ which converges if y< 1.

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