f(x)=x^2+bln(x+1) f(x)=x^2+bln(x+1) (2) if B = 1, we prove that for any positive integer n, the inequality ∑ f (1 / k), 1 + 1 / 2 ^ 3 + 1 / 3 ^ 3 +. + 1 / N ^ 3 I think we should use mathematical induction, but I didn't try it out Is f (1 / 1) + F (1 / 2) + F (1 / 3) +... + F (1 / N)

f(x)=x^2+bln(x+1) f(x)=x^2+bln(x+1) (2) if B = 1, we prove that for any positive integer n, the inequality ∑ f (1 / k), 1 + 1 / 2 ^ 3 + 1 / 3 ^ 3 +. + 1 / N ^ 3 I think we should use mathematical induction, but I didn't try it out Is f (1 / 1) + F (1 / 2) + F (1 / 3) +... + F (1 / N)

If B = 1, it should be,
F (1 / 1) + F (1 / 2) + F (1 / 3) +. + F (1 / N) > 1 + 1 / 2 ^ 3 + 1 / 3 ^ 3 +. + 1 / N ^ 3
Because, on the left
f(1/1)+f(1/2)+f(1/3)+.+f(1/n)
=∑[(1/k)^2+ln((1/k)+1)]
=∑(1/k)^2+∑ln((1/k)+1)
=∑(1/k)^2+ln∏((1/k)+1)
=∑(1/n)^2+ln[((1/1)+1)((1/2)+1)…… ((1/n)+1)]
=∑(1/n)^2+ln[(2/1)(3/2)…… ((1+n)/n)]
=∑1/n^2+ln(1+n)
right
1+1/2^3+1/3^3+ .+1/n^3
=∑1/n^3
Obviously, for any n > 1, 1 / N ^ 2 > 1 / N ^ 3
So, if and only if n = 1,
In this case, ln (1 + n) = LN2 > 0
Therefore, ∑ 1 / N ^ 2 + ln (1 + n) > ∑ 1 / N ^ 3 holds for any positive integer n
You are likely to copy the wrong place, B = 1, here if it is b = - 1, then you require the certificate to be valid
In this case, it is equivalent to seeking
∑1/n^2-ln(1+n)