Ln(n) [ 1+ Ln(1+1/n) /Ln(n) ] = Ln(n)

topsquark Math Team May 2013 2,518 1,049 The Astral plane Jan 20, 2020 #2 SunDrop said: Ln(n) [ 1+ Ln(1+1/n) /Ln(n) ] = Ln(n) Click to expand... What does Ln(n) mean? This equation seems to be a bit scrambled with the Ln(n) canceling out from both sides. Is there a typo somewhere? -Dan Reactions: SunDrop

SunDrop said: Ln(n) [ 1+ Ln(1+1/n) /Ln(n) ] = Ln(n) Click to expand... What does Ln(n) mean? This equation seems to be a bit scrambled with the Ln(n) canceling out from both sides. Is there a typo somewhere? -Dan

SunDrop Jan 2020 3 0 Morocco Jan 20, 2020 #3 it's not an equation it's like the first term is approximately equals to Ln(n) . Just like sinx = x if you know what i mean ?

it's not an equation it's like the first term is approximately equals to Ln(n) . Just like sinx = x if you know what i mean ?

skipjack Forum Staff Dec 2006 21,478 2,470 Jan 20, 2020 #4 By expanding ln(1 + 1/n) in the LHS for large n, one gets ln(n)(1 + (1/n - . . . )/ln(n)), which has leading term ln(n). (The next term would be 1/n.) Reactions: topsquark and SunDrop

By expanding ln(1 + 1/n) in the LHS for large n, one gets ln(n)(1 + (1/n - . . . )/ln(n)), which has leading term ln(n). (The next term would be 1/n.)

SunDrop Jan 2020 3 0 Morocco Jan 20, 2020 #5 topsquark said: What does Ln(n) mean? This equation seems to be a bit scrambled with the Ln(n) canceling out from both sides. Is there a typo somewhere? -Dan Click to expand... skipjack said: By expanding ln(1 + 1/n) in the LHS for large n, one gets ln(n)(1 + (1/n - . . . )/ln(n)), which has leading term ln(n). (The next term would be 1/n.) Click to expand... Thank youuu!

topsquark said: What does Ln(n) mean? This equation seems to be a bit scrambled with the Ln(n) canceling out from both sides. Is there a typo somewhere? -Dan Click to expand... skipjack said: By expanding ln(1 + 1/n) in the LHS for large n, one gets ln(n)(1 + (1/n - . . . )/ln(n)), which has leading term ln(n). (The next term would be 1/n.) Click to expand... Thank youuu!