- #1

paul-martin

- 27

- 0

**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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- Thread starter paul-martin
- Start date

- #1

paul-martin

- 27

- 0

Can anyone explain this simplification?

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

Kindly Paul-MArtin

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- #2

TD

Homework Helper

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[tex]\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)}}[/tex]

- #3

paul-martin

- 27

- 0

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 :(...

- #4

paul-martin

- 27

- 0

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

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

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

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- #5

Manchot

- 473

- 4

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

- #6

HallsofIvy

Science Advisor

Homework Helper

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1. Your picture says [tex]lim_{x->\infty}[/tex] 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 [tex]lim_{k->\infty}\frac{a_{n+1}}{a_n}= y[/tex]

and y< 1 then the infinite series [tex]\Sigma_{k=1}^{\infty}a_n[/tex]

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, [tex]\frac{a_{k+1}}{a_k}< y[/tex] then we can write a_{k+1}< ya_{k}< y^{2}a_{k-1}< ...< y^{k}a_{1} so that the series in dominated by the geometric series [tex]\Sigma_{k=1}^{\infty}a_1y^k[/tex] which converges if y< 1.

2. If the limit you show equals

I

If the limit [tex]lim_{k->\infty}\frac{a_{n+1}}{a_n}= y[/tex]

and y< 1 then the infinite series [tex]\Sigma_{k=1}^{\infty}a_n[/tex]

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, [tex]\frac{a_{k+1}}{a_k}< y[/tex] then we can write a

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