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J. M.'s missing motivation ♦
J. M.'s missing motivation ♦
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Possible bug in finite sum over inverse squares $\sum _$\sum\limits_{i=1}^n \frac{1}{(x (n-i)+i)^2}$

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A little mouse on the pampas
A little mouse on the pampas
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Revisiting the problem Limit of partial sums involving inverse squares I found another difficulty with Sum[]

Consider this sum

s[x_, n_] :=  Sum[ 1/(i + (n - i) x)^2, {i, 1, n}]

Here we assume x > 0, and n Integer > 0.

We have for example

s[2, 10]

(* Out[11]= 2920725891004177/54192375991353600 *)

But considering the symbolic evaluation gives

s[2, n]

(* Out[9]= PolyGamma[1, 1 - 2 n] - PolyGamma[1, 1 - n] *)

This is definitely a wrong result.

Numerically this becomes even more obvious:

Limit[%, n -> 10]

(* Out[10]= -\[Infinity]∞ *)

I would consider this behaviour of Sum[] as a bug.

Revisiting the problem Limit of partial sums involving inverse squares I found another difficulty with Sum[]

Consider this sum

s[x_, n_] :=  Sum[ 1/(i + (n - i) x)^2, {i, 1, n}]

Here we assume x > 0, and n Integer > 0.

We have for example

s[2, 10]

(* Out[11]= 2920725891004177/54192375991353600 *)

But considering the symbolic evaluation gives

s[2, n]

(* Out[9]= PolyGamma[1, 1 - 2 n] - PolyGamma[1, 1 - n] *)

This is definitely a wrong result.

Numerically this becomes even more obvious:

Limit[%, n -> 10]

(* Out[10]= -\[Infinity] *)

I would consider this behaviour of Sum[] as a bug.

Revisiting the problem Limit of partial sums involving inverse squares I found another difficulty with Sum[]

Consider this sum

s[x_, n_] :=  Sum[ 1/(i + (n - i) x)^2, {i, 1, n}]

Here we assume x > 0, and n Integer > 0.

We have for example

s[2, 10]

(* Out[11]= 2920725891004177/54192375991353600 *)

But considering the symbolic evaluation gives

s[2, n]

(* Out[9]= PolyGamma[1, 1 - 2 n] - PolyGamma[1, 1 - n] *)

This is definitely a wrong result.

Numerically this becomes even more obvious:

Limit[%, n -> 10]

(* Out[10]= -∞ *)

I would consider this behaviour of Sum[] as a bug.

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Dr. Wolfgang Hintze
Dr. Wolfgang Hintze
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Possible bug in finite sum over inverse squares $\sum _{i=1}^n \frac{1}{(x (n-i)+i)^2}$

Revisiting the problem Limit of partial sums involving inverse squares I found another difficulty with Sum[]

Consider this sum

s[x_, n_] :=  Sum[ 1/(i + (n - i) x)^2, {i, 1, n}]

Here we assume x > 0, and n Integer > 0.

We have for example

s[2, 10]

(* Out[11]= 2920725891004177/54192375991353600 *)

But considering the symbolic evaluation gives

s[2, n]

(* Out[9]= PolyGamma[1, 1 - 2 n] - PolyGamma[1, 1 - n] *)

This is definitely a wrong result.

Numerically this becomes even more obvious:

Limit[%, n -> 10]

(* Out[10]= -\[Infinity] *)

I would consider this behaviour of Sum[] as a bug.