Question

A sequence of numbers (a_n) is defined as

Answer 1

\[a_1=1/2\] 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 \[n \ge1\]

Answer 2

Just try to Prove

Answer 3

\[a_{1}+ a_{2}+…+ a_{n} < 1\]
\[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2n - 3 }{ 2n })a_{n-1} < 1\]
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

Answer 4

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= \[1 + (\frac{ 2(m+1) - 3 }{ 2(m+1) })a_{m}\]
\[=(\frac{ 2m-1 }{ 2m+2 })(\frac{ 2m - 3 }{ 2m })a_{m-1}\]

Answer 5

@sauravshakya @azteck
thanks

Answer 6

not really worth 1 hour

Answer 7

I am not sure if this helps