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I'm to prove this by mathematical induction: Edited: I made a typo error. $\sum_{k=1} ^n (k^2+1)k!=n(n+1)!$

I made the test and the rightside is true.

So I tested: $N+1$

$N(N+1)! + (N^2+1)N!$

$N(N+1)(N!)+(N^2+1)N!$

$N!\left[N(N+1)+N^2+1\right]$

$N! (N^2+N+N^2+1)$

$\sum_{k=1} ^ {N+1} (k^2+1)k!=$ $N!(2N^2+N+1)$

Is this a valid proof?

Orestes Dante
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    No - because you did not end up with the desired result. Another issue is that you added $(N^2+1)N!$ at the start rather than $(N+1+1),(N+1)!$. And at first sight, when $n=2$ you have $\sum_{k=1} ^n (k+1)k! = 2\times 1+3\times 2=8$ but $n(n+1)!=2\times 6=12$ – Henry Sep 21 '19 at 08:49
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    The statement is false. – Cave Johnson Sep 21 '19 at 09:02
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    What would be true is $\sum\limits_{k=1} ^n k\cdot k!=(n+1)!-1$ – Henry Sep 21 '19 at 09:02
  • Without induction $k^2k!+k!=k(k+1-1)k!+k!=k((k+1)!-k!)+k!=k(k+1)!-(k-1)k!$ and this sum is telescoping. – kingW3 Sep 21 '19 at 09:20
  • https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 – lab bhattacharjee Sep 21 '19 at 10:28

3 Answers3

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There are various problems, including

  • the thing you are trying to prove is not correct, for example when $n=2$
  • you did not end up showing the inductive hypothesis was true for $N+1$
  • you seem to have added $(N^2+1)N!$ rather than $(N+1+1)\,(N+1)!$

So let's illustrate proving something that is true: $$\sum\limits_{k=1} ^n k\cdot k!=(n+1)!-1$$

First we check that it is correct at the start, when $n=1$: we have $1 \cdot 1!=1$ on the left hand side and $2!-1=1$ on the right and these are indeed equal

Then we assume it is true for $n=m$ and consider what happens trying to go forward a step:

$\sum\limits_{k=1} ^{m+1} k\cdot k! = \sum\limits_{k=1} ^m k\cdot k! + (m+1)\cdot (m+1)! \\ =(m+1)!-1+(m+1)\cdot (m+1)! \\= (m+1)!(1+m+1)-1 \\= (m+2)! -1$

which shows it would also be true for $n=m+1$

and therefore, by induction, it is true for all positive integer $n$

Henry
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Your inductive step doesn't work because your expression doesn't simplify to $(N+2)!(N+1)$.

Note that $$\sum_{k=1}^n(k^2+1)k!=\sum_{k=1}^n((k+2)!-3(k+1)!+2\cdot k!)\\=\sum_{k=1}^n((k+2)!-(k+1)!+2\cdot k!-2\cdot (k+1)!)\\=(n+2)!-2\cdot (n+1)!=(n+1)!n.$$

J.G.
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$\sum _{k=1} ^{N+1} (k^2+1)k!$=$[(N+1)^2 +1]!+N(N+1)!\\ =(N+1)![(N^2+3N+2]\\ =(N+1)(N+1)(N+2)\\ =(N+1)(N+2)!$

Orestes Dante
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