United States Calculus 2 - Infinite Series

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
GreenPrint
Messages
1,186
Reaction score
0

Homework Statement



Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10^-5. Although you do not need it, the exact value of the series is given.

ln(128) = 7*sum[k=1,inf] of (-1)^(k+1)/k

Homework Equations





3. The Attempt at a Solution [/b

| ln(128) - 7*sum[k=1,n] of (-1)^(k+1)/k | < 1/10,000
subtracted ln(128) from both sides
|-7*sum[k=1,n] of (-1)^(k+1)/k | < 1/10,000 - ln(128)
simplified the negative on the left hand side and absolute value
7*sum[k=1,n] of 1/k < 1/10,000 - ln(128)
divided through by 7
sum[k=1,n] of 1/k < 1/70,000 - ln(128)/7

I'm unsure were to go from here, thank you for any help you can provide me.
 
Physics news on Phys.org
GreenPrint said:

Homework Statement



Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10^-5. Although you do not need it, the exact value of the series is given.

ln(128) = 7*sum[k=1,inf] of (-1)^(k+1)/k

Homework Equations





3. The Attempt at a Solution [/b

| ln(128) - 7*sum[k=1,n] of (-1)^(k+1)/k | < 1/10,000
subtracted ln(128) from both sides
|-7*sum[k=1,n] of (-1)^(k+1)/k | < 1/10,000 - ln(128)
simplified the negative on the left hand side and absolute value
7*sum[k=1,n] of 1/k < 1/10,000 - ln(128)
divided through by 7
sum[k=1,n] of 1/k < 1/70,000 - ln(128)/7

I'm unsure were to go from here, thank you for any help you can provide me.

There's a remainder theorem for alternating series. Take a look at it.