Prove that Σ(n = 0 to ∞) [1/(an + b)] , where a and b are any real numbers and none of the term of an + b is 0 , is always divergent.
It is so said that http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#General_harmonic_series only state but not prove.
About general harmonic series
2009-01-18 4:59 am
回答 (2)
2009-01-18 5:36 am
✔ 最佳答案
By comparsion with the divergent series Σ(n = 0 to ∞) 1/nlim (n-> ∞) an/bn=lim(n->∞) (1/an+b)/(1/n)=n/(an+b)=1/a>0
From the limit comparsion test, Σ(n = 0 to ∞) [1/(an + b)] , where a and b are any real numbers and none of the term of an + b is 0 , is always divergent
2009-01-30 11:50 pm
Since none of an + b is zero, b is non-zero since if b = 0, an + b = 0 when n = 0
So by case consideration:
Case 1: a,b > 0
http://i117.photobucket.com/albums/o61/billy_hywung/Jan09/Crazyseries7.jpg
And for a,b < 0, similar approach.
2009-01-30 15:50:31 補充：
Case 2: a > 0, b < 0
http://i117.photobucket.com/albums/o61/billy_hywung/Jan09/Crazyseries8.jpg
http://i117.photobucket.com/albums/o61/billy_hywung/Jan09/Crazyseries9.jpg
And for a < 0, b > 0, similar approach
2009-01-30 15:51:36 補充：
Case 3: a = 0
Result is obvious since all terms are of equal non-zero value.
So by case consideration:
Case 1: a,b > 0
http://i117.photobucket.com/albums/o61/billy_hywung/Jan09/Crazyseries7.jpg
And for a,b < 0, similar approach.
2009-01-30 15:50:31 補充：
Case 2: a > 0, b < 0
http://i117.photobucket.com/albums/o61/billy_hywung/Jan09/Crazyseries8.jpg
http://i117.photobucket.com/albums/o61/billy_hywung/Jan09/Crazyseries9.jpg
And for a < 0, b > 0, similar approach
2009-01-30 15:51:36 補充：
Case 3: a = 0
Result is obvious since all terms are of equal non-zero value.
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