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3 solutions
We can wrote n n + 1 = 1 + n 1 and n + 1 n + 2 = 1 + n + 1 1 , hence n n + 1 − n + 1 n + 2 = ( 1 + n 1 ) − ( 1 + n + 1 1 ) = n 1 − n + 1 1 Hence when we put sigma, it will be m → ∞ lim ( n = 1 ∑ m ( n 1 − n + 1 1 ) ) = m → ∞ lim ( 1 − m + 1 1 ) = 1
Indeed. Alternatively you can remove the 1 from every term as Kay did here. Ultimately, it's about accounting for lim m → ∞ T ( m ) .
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It is possible to do it without removing terms as well. If doing so, the limit evaluation will change to l i m m → ∞ n = 1 ∑ m ( 2 − n + 1 n + 2 ) = l i m m → ∞ ( 2 − m + 1 m + 2 ) = 2 − 1 = 1
[This is not a complete solution.]
Most people will think that this is an obvious infinite telescoping series, so all that we need to evaluate is T ( 1 ) = 1 1 + 1 = 2 .
But why isn't this the answer?
There will always be a non-zero term at the end which doesn't cancel.
To see how we shall consider the partial sum which telescopes:
n = 1 ∑ m ( n n + 1 − n + 1 n + 2 ) = 1 1 + 1 − m + 1 m + 2
Now let's take the limit as m tends to ∞ :
m → ∞ lim n = 1 ∑ m ( n n + 1 − n + 1 n + 2 ) = m → ∞ lim [ 1 1 + 1 − m + 1 m + 2 ] = 2 − 1 = 1
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Nice explanation. I nearly chose 2 as the answer until I rewrote the summand as 1 + n 1 − n + 1 ( n + 1 ) + 1 = n 1 − n + 1 1 , which does genuinely telescope.
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My sequence does genuinely telescope. It's just that the telescope is equal to T ( 1 ) − T ( n ) , and so when taking the limit to infinity, we actually get T ( 1 ) − lim n → ∞ T ( n ) . However, many people think that T ( ∞ ) = 0 , which need not be true.
Can you post this directly as a solution? Thanks!
Can we cancel infinity upon infinity
I fell for that. :(
The remained: \frac{1 + 1}{1} -~^\lim_{n \to \infty} \frac{n + 2}{n + 1} = 2 - 1 = 1
Alternative explanation: rewriting the series as the sum 2 = 1 2 + ( − 2 3 + 2 3 ) + ( − 3 4 + 3 4 ) + ⋯ is only valid if the series 1 2 − 2 3 + 2 3 − 3 4 + 3 4 − ⋯ is absolutely convergent; which of course it isn't (by e.g. the alternating series test) :)
n = 1 ∑ ∞ ( n n + 1 − n + 1 n + 2 ) = n = 1 ∑ ∞ n ( n + 1 ) n 2 + 2 n + 1 − n ( n + 1 ) n 2 + 2 n = n = 1 ∑ ∞ n ( n + 1 ) 1
Then decompose the fraction using n ( n + 1 ) 1 = n 1 − n + 1 1
n = 1 ∑ ∞ ( n n + 1 − n + 1 n + 2 ) = n = 1 ∑ ∞ ( n 1 − n + 1 1 )
= n → ∞ lim ( 1 1 − 2 1 + 2 1 − 3 1 + 3 1 − 4 1 + 4 1 + . . . − n 1 + n 1 − n + 1 1 )
= n → ∞ lim ( 1 − n + 1 1 ) = 1 .