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# Show that the limit of ∑ 1/k = log (p/q) where the sum is from k = qn to pn

1. For given integers and with , prove 2. Consider the series This is a rearrangement of the alternating harmonic series ( ) in which there are three positive terms followed by two negative terms. Prove that the series converges and that the sum is equal to .

Incomplete.

1. Evangelos says:

Didn’t even realize there was a part (c) waiting for us on the next page before the start of section 10.23 lol. Here it goes.

(c) Rearrange the alternating harmonic series, writing alternately p positive terms followed by q negative terms. Then, use part (a) to show that this rearranged series converges and has sum .

To show the above, we note that in the alternating harmonic series, every odd term is positive and every even term is negative. Now, consider the partial sum , where m = (p+q)n for some integer n. This gives us the following: Using the results of parts (a) and (b), we get 2. Anonymous says:

Prove it as it was show in the begining of section *10.21.

• Evangelos says:

The patterning of this problem is slightly different than that of the example in 10.21, but the principle is the same.

(a) To show that where p and q are integers, with , we first note that the above limit can be rewritten as follows Then, using the limit relation from example 3, section 10.17 where C is Euler’s constant, we get the following as requested.

(b) To show that the sum of approaches the value as n goes to infinity, we first consider the partial sum, , which can be written as follows Using the results from part (a), as well as the limit relation from example 3, section 10.17, we get the following limit relation • Evangelos says:

Didn’t even realize there was a part (c) waiting for us on the next page before the start of section 10.23 lol. Here it goes.

(c) Rearrange the alternating harmonic series, writing alternately p positive terms followed by q negative terms. Then, use part (a) to show that this rearranged series converges and has sum To show the above, we note that in the alternating harmonic series, every odd term is positive and every even term is negative. Now, consider the partial sum s_m, where m = (p+q)n for some integer n. This gives us the following: Using the results of parts (a) and (b), we get 