9514 1404 393
Answer:
sum = 1
Step-by-step explanation:
The method of partial sums has you look at sums of various numbers of terms of the series to see if there is a pattern that suggests either divergence or a limit. Here, the partial sums suggest the series sum is 1.
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In the attachment, we have computed sums of 1 term, 10 terms, 100 terms, and so on up to 10^8 terms. These sums seem to approach the value 1.
Individual terms get small faster than for a harmonic series, which his known not to converge, but more slowly than a quadratic series. Large numbered terms are approximately proportional to n^-(3/2).
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You can actually compute the sum exactly, without reference to partial sums. Each term of the sum can be rewritten as ...
[tex]\dfrac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})}\\\\=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}[/tex]
Then successive terms will cancel the intermediate fractions. So, for example, the sum of the first 3 terms is ...
(1/√1 -1/√2) +(1/√2 -1/√3) +(1/√3 -1/√4) = 1 -1/√4
This means the given sum will evaluate to ...
1 -1/√∞ = 1 -0 = 1 . . . sum of the infinite series
