MHB Find closed form expression for a given sum

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The discussion revolves around finding a closed form expression for the sum $$\sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}$$. Participants express gratitude for a solution provided by kaliprasad, while also sharing their own similar approaches. One participant acknowledges a mistake in their solution that went unnoticed. The conversation highlights the collaborative effort to solve the mathematical problem. Overall, the focus remains on deriving the correct closed form expression for the given sum.
anemone
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Find a closed form expression for $$\sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}$$.
 
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anemone said:
Find a closed form expression for $$\sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}$$.

each term = $(n- \lfloor \sqrt{k-1}\rfloor)(\sqrt{k+1}-\sqrt{k})$

if we take k fron $m^2+1$ to $(m+1)^2$ that is (2m+1) terms we get

$(n- m)(\sqrt{k+1}-\sqrt{k})$

summing from $m^2+1$ to $(m+1)^2$ we get (n-m)



now we need to add (n-m) with m from 1 to n to get the sum as n^2 - n(n+1)/2 = n(n-1)/2
 
Thank you kaliprasad for your clever solution and sorry for the late reply...:(

I will share with you the quite similar but not completely the same solution with you and MHB:

Note that

$\begin{align*}\displaystyle \sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}&=\sum_{m=1}^{n}\sum_{k=(m-1)^2+1}^{m^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}\sum_{k=(m-1)^2+1}^{m^2}\dfrac{n-(m-1)}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}(n-m+1)\sum_{k=(m-1)^2+1}^{m^2}\dfrac{1}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}(n-m+1)\sum_{k=(m-1)^2+1}^{m^2}\sqrt{k}-\sqrt{k+1}\\&=\sum_{m=1}^{n}(n-m+1)(m-(m-1))\\&=\sum_{m=1}^{n}(n-m+1)\\&=n(n+1)-\dfrac{n(n+1)}{2}\\&=\dfrac{n(n+1)}{2}\end{align*}$
 
anemone said:
Thank you kaliprasad for your clever solution and sorry for the late reply...:(

I will share with you the quite similar but not completely the same solution with you and MHB:

Note that

$\begin{align*}\displaystyle \sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}&=\sum_{m=1}^{n}\sum_{k=(m-1)^2+1}^{m^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}\sum_{k=(m-1)^2+1}^{m^2}\dfrac{n-(m-1)}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}(n-m+1)\sum_{k=(m-1)^2+1}^{m^2}\dfrac{1}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}(n-m+1)\sum_{k=(m-1)^2+1}^{m^2}\sqrt{k}-\sqrt{k+1}\\&=\sum_{m=1}^{n}(n-m+1)(m-(m-1))\\&=\sum_{m=1}^{n}(n-m+1)\\&=n(n+1)-\dfrac{n(n+1)}{2}\\&=\dfrac{n(n+1)}{2}\end{align*}$

Hello Anemone,
Thanks

However In my solution there is a mistake you did not notice

I forgot to add the last value that is n to give n(n+1)/2
 
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