Bois-Reymond criterion for series

  • Thread starter Thread starter mahler1
  • Start date Start date
  • Tags Tags
    Series
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 1K views
mahler1
Messages
217
Reaction score
0
Homework Statement .

Prove that if ##\sum_{n=1}^{\infty} (a_n-a_{n-1})## converges absolutely and ##\sum_{n=1}^{\infty} z_n## converges, then ##\sum_{n=1}^{\infty} a_nz_n## converges.

The attempt at a solution.

I know that if ##Z_N=z_0+z_1+...+z_N##, then ##\sum_{n=0}^N a_nz_n= a_NZ_N-\sum_{n=0}^{N-1} Z_n(a_{n+1}-a_n)##

I am not so sure how can I use the hypothesis given to this new expression or if it would be more convenient to express the original series in another way.
 
Last edited:
Physics news on Phys.org
Your formula ##\sum_{n=0}^N a_nz_n= a_NZ_N-\sum_{n=0}^{N-1} Z_N(a_{n+1}-a_n)## is incorrect.

##a_NZ_N-\sum_{n=0}^{N-1} Z_N(a_{n+1}-a_n)=0##.
 
  • Like
Likes   Reactions: 1 person
xiavatar said:
Your formula ##\sum_{n=0}^N a_nz_n= a_NZ_N-\sum_{n=0}^{N-1} Z_N(a_{n+1}-a_n)## is incorrect.

##a_NZ_N-\sum_{n=0}^{N-1} Z_N(a_{n+1}-a_n)=0##.

I've corrected it