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## Jonask 2 years ago A sequence of numbers (a_n) is defined as

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1. Jonask

$a_1=1/2$ and for each $n \ge 2$ $\huge a_n=\left(\frac{ 2n-3 }{ 2n }\right)a_{n-1}$ Prove that $\huge\sum_{k=1}^{n}a_k<1$ for all $n \ge1$

2. sauravshakya

|dw:1357738064697:dw|Just try to Prove

3. Azteck

$a_{1}+ a_{2}+…+ a_{n} < 1$ $\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2n - 3 }{ 2n })a_{n-1} < 1$ For n=1 LHS=$\frac{ 1 }{ 2 }$ $<RHS$ (Therefore) True for n=1 For n=2 LHS=$(\frac{ 2(2) - 3 }{ 2(2) })a_{2-1}$ $<RHS$ (Therefore) True for n=2

4. Azteck

Assume the formula is true for n=m $\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2m - 3 }{ 2m })a_{m-1} < 1$ For n=m+1 $\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2m - 3 }{ 2m })a_{m-1}+(\frac{ 2(m+1) - 3 }{ 2(m+1) })a_{m+1-1} < 1$ LHS= $1 + (\frac{ 2(m+1) - 3 }{ 2(m+1) })a_{m}$ $=(\frac{ 2m-1 }{ 2m+2 })(\frac{ 2m - 3 }{ 2m })a_{m-1}$

5. Jonask

@sauravshakya @azteck thanks

6. Jonask

not really worth 1 hour

7. sauravshakya

I am not sure if this helps

8. sauravshakya

|dw:1359039410279:dw|

9. sauravshakya

Did u mean like this @experimentX

10. experimentX

yeah .. somewhat similar to that. but one terms is missing.

11. experimentX

there should be one n ... or i made mistake.

12. sauravshakya

|dw:1359040853679:dw|

13. sauravshakya

It was me who did a mistake

14. experimentX

hmm ... now lets use this identity. http://planetmath.org/GeneratingFunctionForTheCatalanNumbers.html

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