Infinite Series: U(n+1)/U(n) Calculation

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rohit dutta
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The given series is:
1+[(a+1)/(b+1)]+[(a+1)(2*a+1)/(b+1)(2*b+1)]+[(a+1)(2*a+1)(3*a+1)/(b+1)(2*b+1)(3*b+1)]+...∞

Problem:
To find U(n+1)/U(n).

My approach:

Removing the first term(1) of the series and making the second term the first,third term the second and so on...
I get,
U(n+1)/U(n)={(n+1)a}+1/{(n+1)b}+1.

Text book approach:

Neglecting the first term and keeping the position of the succeeding terms unchanged, it gives,
U(n+1)/U(n)={n*a}+1/{n*b}+1.

I believe my approach is right because solving further to test for convergence or divergence, both of us end up with the same answer. Also, according to the property, removing or adding a term will not affect the convergence or divergence of a series. But, in my approach, I changed the positions of the terms after the removal of the first term. It did not affect my answer but is it the right way to solve the problem? Will my approach always assure a right answer?
 
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hi rohit dutta! :smile:
rohit dutta said:
I believe my approach is right because solving further to test for convergence or divergence, both of us end up with the same answer. Also, according to the property, removing or adding a term will not affect the convergence or divergence of a series. But, in my approach, I changed the positions of the terms after the removal of the first term. It did not affect my answer but is it the right way to solve the problem? Will my approach always assure a right answer?

yes, your approach will always give the right answer

however, in this case the book's approach is better, for two reasons …

i] it gives factors of n (instead of n+1), which is simpler
ii] the less re-numbering you do the less likely you are to make a mistake! :wink:
 
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Mark44 said:
Maybe I'm dense, but what are U(n) and U(n + 1)?

U(n) appears to be the first n terms of …
rohit dutta said:
The given series is:
1+[(a+1)/(b+1)]+[(a+1)(2*a+1)/(b+1)(2*b+1)]+[(a+1)(2*a+1)(3*a+1)/(b+1)(2*b+1)(3*b+1)]+...∞

:wink:
 
U(n) and U(n+1) refer to the nth and (n+1)th term of the series respectively.