Could you find the result of the attached file (Wolfram Alpha Pro can’t because time of calculation is too long)?
If you think very carefully then can you justify rewriting your problem as
Sum(Ln(k+2)/(2(k+2))-Ln(k+1)/(2(k+1)),{k,0,n})
Writing your problem in that form means WA will give you an answer in a faction of a second.
Then think very very carefully about what happens with that result as n becomes very very large.
Depending on how much background you have, you might be able to use WA to determine what that result will be as n goes to infinity.
In fact, the two terms you input do not converge. So, if you enter the expression in that form, the program will meet two infinities and can’t add them up.
In addition, if a limit \underset{n \rightarrow \infty}{\lim} a_n - b_n exists (converges), it doesn’t mean series \underset{n\rightarrow \infty}{\lim}a_n and \underset{n\rightarrow \infty}{\lim}b_n both exist (converge).
The two \infty in you expression may be different precisely. The expression you input is equal to
\underset{n\rightarrow \infty}{\lim}\sum_{k=0}^{n} \frac{ln(k+2)}{2(k+2)} - \underset{m\rightarrow \infty}{\lim}\sum_{k=0}^{m}\frac{ln(k+1)}{2(k+1)}
where two series both don’t converge actually. So the sum of them don’t converge.
But if you sum each term together like the reply before, you will get a series that converge and is easy to calculate the result.
\underset{n\rightarrow \infty}{\lim}(\sum_{k=0}^{n} \frac{ln(k+2)}{2(k+2)} - \sum_{k=0}^{n}\frac{ln(k+1)}{2(k+1)}) = \underset{n\rightarrow \infty}{\lim}\sum_{k=0}^{n}( \frac{ln(k+2)}{2(k+2)} - \frac{ln(k+1)}{2(k+1)})
Here the two \infty are the same and we can add each term of two series up. The series above converge and is equal to 0.
Addition: the two infinite series are different actually. The first one doesn’t converge but the second one converges to a finite value.
Well, this is not MMA-related - and really trivial:
The sums are of the form:
\sum_{k=0}a_{k+2}-\sum_{k=0}a_{k+1}
which gives
\sum_{k=2}a_{k}-\sum_{k=1}a_{k} = \sum_{k=2}a_{k}-\left(a_1+\sum_{k=2}a_{k}\right) = -a_1
In this case we have
a_k=\frac{\ln k}{2k} \quad\Longrightarrow\quad a_1=0
EDIT: OK, after some thinking I guess in a strong mathematical sense my approach is not really correct, because effectively I am doing a term wise cancellation - but in infinite sums the commutative law does not hold …
Many thanks!
Many thanks Bill!
Many thanks!