Power series problem !!

[ronjan]

My 1st thread...because am badly stuck !!!

Does this series have a simple sum ?

x^(a^1) + x^(a^2) + x^(a^3) + x^(a^4) + .... + x^(a^n)

where |x| < 1 and |a| < 1 .

Thanx
ranjan

[Aarav]

May be try to differentiate and then integrate the sum(as a function of a). I've forgotten the formulas.

something like this comes...on differentiating

log(x) *(x^a + 2x^2a + 3x^3a + 4x^4a...), i think very much solvable if you can remember calculus.

[khanna_sumit]

Originally Posted by ronjan

My 1st thread...because am badly stuck !!!

Does this series have a simple sum ?

x^(a^1) + x^(a^2) + x^(a^3) + x^(a^4) + .... + x^(a^n)

where |x| < 1 and |a| < 1 .

Thanx
ranjan

this series has no sum because it is divergent

lets compare x^a with x^a^2

x^a^2=(x^a)^a since x^a is already a fraction and so is a therefore x^a raised to poer a will be greater that x^a ....u can check that by taking any example......since each subsequent term is greater that the previous term there fore sum cannot be found...

[ronjan]

Originally Posted by khanna_sumit

this series has no sum because it is divergent

lets compare x^a with x^a^2

x^a^2=(x^a)^a since x^a is already a fraction and so is a therefore x^a raised to poer a will be greater that x^a ....u can check that by taking any example......since each subsequent term is greater that the previous term there fore sum cannot be found...

But each term is <1 and hence the upper bound on any n-term series is 'n', infact a tighter bound can be found which would hold for n-> infinity.

Hence, I dont think the series is divergent.

[Eccentric]

Originally Posted by ronjan

But each term is <1 and hence the upper bound on any n-term series is 'n', infact a tighter bound can be found which would hold for n-> infinity.

Hence, I dont think the series is divergent.

sumit: even the series 1 1/2 1/3 1/4 ... is divergent where we have an upper bound of 1/n and we can't find the sum.

[khanna_sumit]

Originally Posted by Eccentric

sumit: even the series 1 1/2 1/3 1/4 ... is divergent where we have an upper bound of 1/n and we can't find the sum.

i believe this logic should be correct