siriusmart@lemmy.worldM to Daily Maths Challenges@lemmy.world · 6 months ago[2024/05/07] Evaluate converging sumlemmy.worldimagemessage-square3fedilinkarrow-up119arrow-down12file-text
arrow-up117arrow-down1image[2024/05/07] Evaluate converging sumlemmy.worldsiriusmart@lemmy.worldM to Daily Maths Challenges@lemmy.world · 6 months agomessage-square3fedilinkfile-text
minus-squarezkfcfbzr@lemmy.worldlinkfedilinkEnglisharrow-up4·6 months ago solution With partial fractions: 1/(n + n²) = 1/(n(n+1)) = A/n + B/(n+1) A(n+1) + Bn = 1 n = 0 gives A = 1, n = -1 gives B = -1 1/(n+n²) = 1/n - 1/(n+1) Σ (n = 1 to ∞) 1/(n+n²) = Σ (n = 1 to ∞) 1/n - Σ (n = 1 to ∞) 1/(n+1) = Σ (n = 1 to ∞) 1/n - Σ (n = 2 to ∞) 1/n = 1/1 + Σ (n = 2 to ∞) 1/n - Σ (n = 2 to ∞) 1/n = 1 Guessing this is the standard solution
solution
With partial fractions:
1/(n + n²) = 1/(n(n+1)) = A/n + B/(n+1)
A(n+1) + Bn = 1
n = 0 gives A = 1, n = -1 gives B = -1
1/(n+n²) = 1/n - 1/(n+1)
Σ (n = 1 to ∞) 1/(n+n²) = Σ (n = 1 to ∞) 1/n - Σ (n = 1 to ∞) 1/(n+1)
= Σ (n = 1 to ∞) 1/n - Σ (n = 2 to ∞) 1/n
= 1/1 + Σ (n = 2 to ∞) 1/n - Σ (n = 2 to ∞) 1/n
= 1
Guessing this is the standard solution