In previous
entries, Interesting
Log Relationships (1) and Interesting
Log Relationships (2), I showed how the (infinite) reciprocal sum of the
unique number sequences associated with (x – k)

(x – k)

^{2}= 0 and(x – k)

^{3 }= 0, (with k = 2, 3, 4, …) together with (x + k)^{2}= 0 and (x + k)^{3 }= 0 (with k = 1, 2, 3, …), result from the application of generally applicable formulae resulting in answers of the form a log k + b (where a and b are rational numbers that can take on positive or negative values).
So once
again for example for the general case where (x + k)

^{3 }= 0, the general formula is
2k{(k + 1)[log(k + 1/k)] – 1}.

Thus when k = 1, we have the polynomial equation (x + 1)

^{3 }= 0 i.e. x^{3 }+ 3x^{2}+ 3x + 1 = 0, with the unique number sequence 1, – 3, 6, – 10, 15, …
Therefore the corresponding (infinite) sum of reciprocals
(i.e. the alternating sum of reciprocals of the triangular nos.) = 1 – 1/3 +
1/6 – 1/10 + 1/15 …

= 2{2(log 2) – 1} = 4 log 2 – 2.

However one can continue on both for higher powers of (x – k)

^{n}= 0 and (x + k)^{n }= 0, to generate corresponding results for the associated (infinite) sums of reciprocals relating to the unique number sequences occurring.
Now while
it appears that the general form of these results does not change (remaining as a log k + b,
it is not yet clear to me whether for any given power of n, that a general
formula expression can be given to cover all possible results.

However, I
did manage to calculate some specific results (which, not surprisingly, become
somewhat more difficult to establish as the power of n and value of k increases).

For example
in the simplest case where (x – k)

^{2}= 0 (with k = 2), we have (x – 2)^{4}= 0,
i.e. x

^{4}– 8x^{3}+ 24x^{2 }– 32x + 16 = 0 with the corresponding unique number sequence
1, 8. 40, 160, 560, … (A001789
in OEIS).

And the corresponding (infinite) sum of reciprocals is,

1 + 1/8 + 1/40 + 1/160 + 1/560 + …, = 1.15888 …
= 6 log 2 – 3.

In the next case where (x – 3)

^{4}= 0, i.e. x^{4}– 12x^{3}+ 54x^{2 }– 108x + 81 = 0, the unique numbers associated are 1, 12, 90, 540, 2835, … (A036216 in OEIS).
Thus the corresponding (infinite) sum of reciprocals is,

1 + 1/12 + 1/90 +
1/540 + 1/2835 + …, = 1.09674… = {72
log (3/2) – 27}/2.

Then in the next case, where (x – 4)

^{4}= 0, i.e. x^{4}– 16x^{3}+ 96x^{2 }– 256x + 256 = 0,
the unique associated number sequence is

1, 16, 160, 1280, 8960, … (i.e. A038846 in OEIS).

Thus the corresponding sum of reciprocals

= 1 + 1/16 + 1/160 + 1/1280 + 1/8960 + … = 1.06966…
= 108 log (4/3) – 30

Then in the case where (x + 1)

^{4}= 0, i.e. x

^{4}+ 4x

^{3}+ 6x

^{2 }+ 4x + 1 = 0, the unique associated number sequence is 1, – 4, 10, – 20, 35, …

So the corresponding (infinite) sum of reciprocals (i.e. sum
of alternating reciprocal terms of the tetrahedral numbers) is,

1 – 1/4 + 1/10 – 1/20 + 1/35, …, = .817766… = (24 log 2 – 15)/2.

Then in the
case where (x + 2)

^{4}= 0, i.e. x^{4}+ 8x^{3}+ 24x^{2 }+ 32x + 16 = 0, the unique associated number sequence is 1, – 8, 40, – 160, 560, …
So the corresponding (infinite) alternating sum of
reciprocals,

= 1 – 1/8 + 1/40 – 1/160 + 1/560 – … = .89511…, = 54 log (3/2) – 21.

And in the case where (x – 5)

So the sum of reciprocals = 1 + 1/20 + 1/250 + 1/2500 + ... , (A081143 in OEIS).

= {480 log (5/4) – 105)}/2.

And in the case where (x – 5)

^{4 }= 0 i.e. x^{4}– 20x^{3}+ 150x^{2 }– 500x + 625 = 0, the unique digit sequence is 0, 0, 0, 1, 20, 250, 2500, ...So the sum of reciprocals = 1 + 1/20 + 1/250 + 1/2500 + ... , (A081143 in OEIS).

= {480 log (5/4) – 105)}/2.

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