Find the value of
1 + 1 + 1 + 1 + ⋱ 1 1 1 1
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
FYI You do not need to restrict to "positive" value. Do you know why?
Log in to reply
Can you explain me clearly.
Log in to reply
If we take negative sign then the answer would be negative but 1+1÷(1+1÷(1× ....)...) Is positive
Log in to reply
@Genis Dude – I got it! cool!
@Genis Dude – Not just that.
Note that the implication sign is only in 1 direction. What this means is that we only have a necessary condition - Any limit must satisfy the given equation. This condition is not sufficient, meaning that any solution to the given equation may not be the limit.
Thus, you're still missing the step of "Justifying that we indeed have found the limit", as opposed to "Well, we only have 1 possible candidate left ...". So, there are two ways to proceed.
1. Showing that
2
1
+
5
is indeed the limit.
2. Prove that the limit exists, and since we only have 1 candidate, that candidate must be the limit.
I just don't understand how this is calculus.
Log in to reply
See my comments. The partial fraction is defined as the limit of truncated terms.
Yes, you can use algebra to solve for it, as Nikhil done above. However, you've not determined the value rigorously. There are instances where the algebraic solution has to be rejected, and we conclude that "the limit does not exist".
Log in to reply
Oh, interesting! I didn't think about it that way. Thanks for showing me!
Log in to reply
@Zach Abueg – As an example of such a scenario, try this sequence. 1 , 2 , 3 .
Log in to reply
@Calvin Lin – Makes sense; those definitely enlighten how I see the concept now. Thanks!
2 1 + 5 = 1 . 6 1 0 8 3 .
Problem Loading...
Note Loading...
Set Loading...
Let, x = 1 + 1 + 1 + 1 + … 1 1 1 1 ⇒ x = 1 + x 1 ⇒ x = x x + 1 ⇒ x 2 = x + 1 ⇒ x 2 − x − 1 = 0 ∴ x = 2 1 + 5 = 1 . 6 1 8 0 3